-
Notifications
You must be signed in to change notification settings - Fork 0
Expand file tree
/
Copy pathlambda.tex
More file actions
executable file
·16398 lines (15017 loc) · 842 KB
/
lambda.tex
File metadata and controls
executable file
·16398 lines (15017 loc) · 842 KB
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
136
137
138
139
140
141
142
143
144
145
146
147
148
149
150
151
152
153
154
155
156
157
158
159
160
161
162
163
164
165
166
167
168
169
170
171
172
173
174
175
176
177
178
179
180
181
182
183
184
185
186
187
188
189
190
191
192
193
194
195
196
197
198
199
200
201
202
203
204
205
206
207
208
209
210
211
212
213
214
215
216
217
218
219
220
221
222
223
224
225
226
227
228
229
230
231
232
233
234
235
236
237
238
239
240
241
242
243
244
245
246
247
248
249
250
251
252
253
254
255
256
257
258
259
260
261
262
263
264
265
266
267
268
269
270
271
272
273
274
275
276
277
278
279
280
281
282
283
284
285
286
287
288
289
290
291
292
293
294
295
296
297
298
299
300
301
302
303
304
305
306
307
308
309
310
311
312
313
314
315
316
317
318
319
320
321
322
323
324
325
326
327
328
329
330
331
332
333
334
335
336
337
338
339
340
341
342
343
344
345
346
347
348
349
350
351
352
353
354
355
356
357
358
359
360
361
362
363
364
365
366
367
368
369
370
371
372
373
374
375
376
377
378
379
380
381
382
383
384
385
386
387
388
389
390
391
392
393
394
395
396
397
398
399
400
401
402
403
404
405
406
407
408
409
410
411
412
413
414
415
416
417
418
419
420
421
422
423
424
425
426
427
428
429
430
431
432
433
434
435
436
437
438
439
440
441
442
443
444
445
446
447
448
449
450
451
452
453
454
455
456
457
458
459
460
461
462
463
464
465
466
467
468
469
470
471
472
473
474
475
476
477
478
479
480
481
482
483
484
485
486
487
488
489
490
491
492
493
494
495
496
497
498
499
500
501
502
503
504
505
506
507
508
509
510
511
512
513
514
515
516
517
518
519
520
521
522
523
524
525
526
527
528
529
530
531
532
533
534
535
536
537
538
539
540
541
542
543
544
545
546
547
548
549
550
551
552
553
554
555
556
557
558
559
560
561
562
563
564
565
566
567
568
569
570
571
572
573
574
575
576
577
578
579
580
581
582
583
584
585
586
587
588
589
590
591
592
593
594
595
596
597
598
599
600
601
602
603
604
605
606
607
608
609
610
611
612
613
614
615
616
617
618
619
620
621
622
623
624
625
626
627
628
629
630
631
632
633
634
635
636
637
638
639
640
641
642
643
644
645
646
647
648
649
650
651
652
653
654
655
656
657
658
659
660
661
662
663
664
665
666
667
668
669
670
671
672
673
674
675
676
677
678
679
680
681
682
683
684
685
686
687
688
689
690
691
692
693
694
695
696
697
698
699
700
701
702
703
704
705
706
707
708
709
710
711
712
713
714
715
716
717
718
719
720
721
722
723
724
725
726
727
728
729
730
731
732
733
734
735
736
737
738
739
740
741
742
743
744
745
746
747
748
749
750
751
752
753
754
755
756
757
758
759
760
761
762
763
764
765
766
767
768
769
770
771
772
773
774
775
776
777
778
779
780
781
782
783
784
785
786
787
788
789
790
791
792
793
794
795
796
797
798
799
800
801
802
803
804
805
806
807
808
809
810
811
812
813
814
815
816
817
818
819
820
821
822
823
824
825
826
827
828
829
830
831
832
833
834
835
836
837
838
839
840
841
842
843
844
845
846
847
848
849
850
851
852
853
854
855
856
857
858
859
860
861
862
863
864
865
866
867
868
869
870
871
872
873
874
875
876
877
878
879
880
881
882
883
884
885
886
887
888
889
890
891
892
893
894
895
896
897
898
899
900
901
902
903
904
905
906
907
908
909
910
911
912
913
914
915
916
917
918
919
920
921
922
923
924
925
926
927
928
929
930
931
932
933
934
935
936
937
938
939
940
941
942
943
944
945
946
947
948
949
950
951
952
953
954
955
956
957
958
959
960
961
962
963
964
965
966
967
968
969
970
971
972
973
974
975
976
977
978
979
980
981
982
983
984
985
986
987
988
989
990
991
992
993
994
995
996
997
998
999
1000
\documentclass[numbers=enddot,12pt,final,onecolumn,notitlepage]{scrartcl}%
\usepackage[all,cmtip]{xy}
\usepackage{lscape}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{amsmath}
\usepackage{amsthm}
\usepackage{hyperref}
\usepackage{comment}
%TCIDATA{OutputFilter=latex2.dll}
%TCIDATA{Version=5.50.0.2960}
%TCIDATA{LastRevised=Monday, November 20, 2017 19:13:24}
%TCIDATA{SuppressPackageManagement}
%TCIDATA{<META NAME="GraphicsSave" CONTENT="32">}
%TCIDATA{<META NAME="SaveForMode" CONTENT="1">}
%TCIDATA{BibliographyScheme=Manual}
%BeginMSIPreambleData
\providecommand{\U}[1]{\protect\rule{.1in}{.1in}}
%EndMSIPreambleData
\iffalse
\newenvironment{proof}[1][Proof]{\noindent\textbf{#1.} }{\ \rule{0.5em}{0.5em}}
\fi
\voffset=-0.5cm
\hoffset=-1.0cm
\setlength\textheight{24cm}
\setlength\textwidth{15.5cm}
\newenvironment{verlong}{}{}
\newenvironment{vershort}{}{}
\newenvironment{noncompile}{}{}
\excludecomment{verlong}
\includecomment{vershort}
\excludecomment{noncompile}
\begin{document}
\title{\fbox{$\lambda$\textbf{-rings: Definitions and basic properties}}}
\author{Darij Grinberg}
\date{Version 0.0.21, last revised
%TCIMACRO{\TeXButton{today}{\today}}%
%BeginExpansion
\today
%EndExpansion
}
\maketitle
\tableofcontents
%\begin{titlepage}
%$\ $\\[20mm]
%\begin{center}
%\textbf{\LARGE $\lambda$\textbf{-rings: Definitions and basic properties}}\\[15mm]
%\Large
%\textit{Darij Grinberg}
%\\[8mm]
%Version ***
%\end{center}
%\end{titlepage}
%\newpage
%$\ \ \ $
%\newpage
\bigskip
This is a \textbf{BETA VERSION} and has never been systematically proofread.
\textbf{Please notify me of any mistakes, typos and hard-to-understand
arguments you find!\footnote{my email address is \texttt{A@B.com}, where
\texttt{A=darijgrinberg} and \texttt{B=gmail}}}
Thanks to Martin Brandenburg for pointing out several flaws.
At the moment, section 1 is missing a proof (namely, that the representation
ring is a \textit{special} $\lambda$-ring; I actually don't know this proof).
Most exercises have solutions or at least hints given at the end of this text;
however, some do not.
\bigskip
\subsection*{What is this?}
These notes try to cover some of the most important properties of $\lambda
$-rings with proofs.
They were originally meant to accompany a talk at an undergraduate seminar,
but quickly grew out of proportion to what could fit into a talk. Still they
lack in anything really deep. At the moment, most of what is written here,
except for the Todd homomorphism section, is also in Knutson's book
\cite{Knut73}, albeit sometimes with different proofs. Part of the plan was to
add some results from the Fulton/Lang book \cite{FulLan85} with better proofs,
but this is not currently my short-term objective, given that I don't
understand much of \cite{FulLan85} to begin with. Most of the notes were
written independently of Yau's 2010 text \cite{Yau10}, but inevitably
intersect with it.
I do not introduce, nor use, the $\lambda$-ring of symmetric functions (see
\cite{Knut73} and \cite[\S 9, \S 16]{Hazewi08b} for it). My avoidance of
symmetric functions has no good reason\footnote{Actually, the reason is that I
have started writing these notes before I understood symmetric functions
well.}; unfortunately, it makes part of the notes (particularly, everything
related to the $\lambda$-verification principle) unnecessarily unwieldy. This
is one of the things I would have done differently if I were to rewrite these
notes from scratch.
\section*{0. Notation and conventions}
Some notations that we will use later on:
\begin{itemize}
\item In the following, $\mathbb{N}$ will denote the set $\left\{
0,1,2,...\right\} $. The elements of this set $\mathbb{N}$ will be called the
\textit{natural numbers}.
\item When we say ``ring'', we will always mean ``commutative ring with
unity''. A ``ring homomorphism'' is always supposed to send $1$ to $1$. When
we say ``$R$-algebra'' (with $R$ a ring), we will always mean ``commutative
$R$-algebra with unity''.
\item Let $R$ be a ring. An \textit{extension ring} of $R$ will mean a ring
$S$ along with a ring monomorphism $R\rightarrow S$. We will often sloppily
identify $R$ with a subring of $S$ if $S$ is an extension ring of $R$; we will
then also identify the polynomial ring $R\left[ T\right] $ with a subring of
the polynomial ring $S\left[ T\right] $, and so on. An extension ring $S$ of
$R$ is called \textit{finite-free} if and only if the $R$-module $S$ is
finite-free (i. e., a free $R$-module with a finite basis).
\item We will use multisets. If $I$ is a set, and $u_{i}$ is an object for
every $i\in I$, then we let $\left[ u_{i}\mid i\in I\right] $ denote the
multiset formed by all the $u_{i}$ where $i$ ranges over $I$ (this multiset
will contain each object $o$ as often as it appears as an $u_{i}$ for some
$i\in I$).\newline If $I=\left\{ 1,2,...,n\right\} $ for some $n\in
\mathbb{N}$, then we also denote the multiset $\left[ u_{i}\mid i\in
I\right] $ by $\left[ u_{1},u_{2},...,u_{n}\right] $.
\item We have not defined $\lambda$-rings yet, but it is important to mention
some discrepancy in notation between different sources. Namely, some of the
literature (including \cite{Knut73}, \cite{Hazewi08a}, \cite{Hazewi08b} and
\cite{Yau10}) denotes as \textit{pre-}$\lambda$\textit{-rings} what we call
$\lambda$-rings and denotes as $\lambda$\textit{-rings} what we call special
$\lambda$-rings. Even worse, the notations in \cite{FulLan85} are totally
inconsistent\footnote{Often, \textquotedblleft$\lambda$-ring\textquotedblright%
\ in \cite{FulLan85} means \textquotedblleft$\lambda$-ring with a positive
structure\textquotedblright\ (such $\lambda$-rings are automatically special),
but sometimes it simply means \textquotedblleft$\lambda$%
-ring\textquotedblright.}.
\item When we say \textquotedblleft monoid\textquotedblright, we always mean a
monoid with a neutral element. (The analogous notion without a neutral element
is called \textquotedblleft semigroup\textquotedblright.) \textquotedblleft
Monoid homomorphisms\textquotedblright\ have to send the neutral element of
the domain to the neutral element of the target.
\item Most times you read an expression with a $\sum$ or a $\prod$ sign in
mathematical literature, you know clearly what it means (e. g., the expression
$\prod\limits_{k=1}^{n}\sin k$ means the product $\left( \sin1\right)
\cdot\left( \sin2\right) \cdot...\cdot\left( \sin n\right) $). However,
some more complicated expressions with $\sum$ and $\prod$ signs can be
ambiguous, like the expression $\prod\limits_{k=1}^{n}\sin k\cdot n$: Does
this expression mean $\left( \prod\limits_{k=1}^{n}\sin k\right) \cdot n$ or
$\prod\limits_{k=1}^{n}\left( \left( \sin k\right) \cdot n\right) $ ? The
answer depends on the author of the text.\newline In \textit{this} text, the
following convention should be resorted to when parsing an expression with
$\sum$ or $\prod$ signs:\newline The argument of a $\prod$ sign ends as early
as reasonably possible. Here, \textquotedblleft reasonably
possible\textquotedblright\ means that it cannot end before the last time the
index of the product appears (e. g., the argument of $\prod\limits_{k=1}%
^{n}\sin k\cdot n$ cannot end before the last appearance of $k$), that it
cannot end inside a bracket (e. g., the argument of $\prod\limits_{k=1}%
^{n}\left( \left( \sin k\right) \cdot n\right) $ cannot end before the end
of the $n$), that it cannot end between a symbol and its exponent or index or
between a function symbol or its arguments, and that the usual rules of
precedence have to apply. For example, the expression $\prod\limits_{k=1}%
^{n}\sin k\cdot n$ has to be read as $\left( \prod\limits_{k=1}^{n}\sin
k\right) \cdot n$, and the expression $\prod\limits_{k=1}^{n}\sin
k\cdot\left( \cos k\right) ^{2}k\cdot\left( n+1\right) kn$ has to be read
as $\left( \prod\limits_{k=1}^{n}\left( \sin k\cdot\left( \cos k\right)
^{2}\cdot\left( n+1\right) k\right) \right) n$.\newline Similar rules
apply to the parsing of a sum expression.
\item Let $R$ be a ring. Let $P\in R\left[ X_{1},X_{2},\ldots,X_{m}%
,Y_{1},Y_{2},\ldots,Y_{n}\right] $ be a polynomial over $R$ in $m+n$
variables. Then, the \textit{total degree} of $P$ with respect to the
variables $X_{1},X_{2},\ldots,X_{m}$ is defined as the highest $d\in
\mathbb{N}$ such that at least one monomial $X_{1}^{a_{1}}X_{2}^{a_{2}}\cdots
X_{m}^{a_{m}}Y_{1}^{b_{1}}Y_{2}^{b_{2}}\cdots Y_{n}^{b_{n}}$ with $a_{1}%
+a_{2}+\cdots+a_{m}=d$ appears in $P$ with a nonzero coefficient. (This total
degree is defined to be $-\infty$ if $P=0$.) Similarly, the total degree of
$P$ with respect to the variables $Y_{1},Y_{2},\ldots,Y_{n}$ is defined.
\item The similarly-looking symbols $\Lambda$ (a capital Lambda) and $\wedge$
(a wedge symbol, commonly used for the logical operator \textquotedblleft
and\textquotedblright) will have completely different meanings. The notation
$\wedge^{i}V$ (where $R$ is a ring, $V$ is an $R$-module and $i$ is a
nonnegative integer) will stand for the $i$-th exterior power of the
$R$-module $V$. On the other hand, the notation $\Lambda\left( K\right) $
(where $K$ is a ring) will stand for a certain ring defined in Chapter 4; this
ring is \textbf{not} the exterior algebra of $K$ (despite some authors
denoting the latter by $\Lambda\left( K\right) $).
\end{itemize}
\section{Motivations}
What is the point of $\lambda$-rings?
Fulton/Lang \cite{FulLan85} motivate $\lambda$-rings through vector bundles.
Here we are going for a more elementary motivation, namely through
representation rings in group representation theory:
\subsection{Representation rings of groups}
Consider a finite group $G$ and a field $k$ of characteristic $0$. In
representation theory, one define the so-called \textit{representation ring}
of the group $G$ over the field $k$. This ring can be constructed as follows:
We consider only finite-dimensional representations of $G$.
Let $\operatorname*{Rep}_{k}G$ be the set of all representations of the group
$G$ over the field $k$. (We disregard the set-theoretic problematics stemming
from the notion of such a big set. If you wish, you can call it a class or a
SET instead of a set, or restrict yourself to a smaller subset containing
every representation up to isomorphism.)
Let $\operatorname*{FRep}_{k}G$ be the free abelian group on the set
$\operatorname*{Rep}_{k}G$. Let $I$ be the subgroup%
\begin{align*}
I & =\left\langle U-V\ \mid\ U\text{ and }V\text{ are two isomorphic
representations of }G\right\rangle \\
& \ \ \ \ \ \ \ \ \ \ +\left\langle U\oplus V-U-V\ \mid\ U\text{ and }V\text{
are two representations of }G\right\rangle
\end{align*}
of the free abelian group $\operatorname*{FRep}_{k}G$ (written additively).
Then, $\operatorname*{FRep}_{k}G\diagup I$ is an abelian group. Whenever $U$
is a representation of $G$, we should denote the equivalence class of
$U\in\operatorname*{FRep}_{k}G$ modulo the ideal $I$ by $\overline{U}$;
however, since we are going to work in $\operatorname*{FRep}_{k}G\diagup I$
throughout this Section 1 (because there is not much of interest to do in
$\operatorname*{FRep}_{k}G$ itself), we will simply write $U$ for this
equivalence class. This means that whenever $U$ and $V$ are two isomorphic
representations of $G$, we will simply write $U=V$, and whenever $U$ and $V$
are two representations of $G$, we will simply write $U+V=U\oplus V$.
Denote by $1$ the equivalence class of the trivial representation of $G$ on
$k$ (with every element of $G$ acting as identity) modulo $I$. We now define a
ring structure on $\operatorname*{FRep}_{k}G\diagup I$ by letting $1$ be the
one of this ring, and defining the product of two representations of $G$ as
their tensor product (over $k$). This is indeed a ring structure because we
have isomorphisms%
\begin{align*}
U\otimes\left( V\otimes W\right) & \cong\left( U\otimes V\right) \otimes
W,\\
\left( U\oplus V\right) \otimes W & \cong\left( U\otimes W\right)
\oplus\left( V\otimes W\right) ,\\
U\otimes\left( V\oplus W\right) & \cong\left( U\otimes V\right)
\oplus\left( U\otimes W\right) ,\\
U\otimes V & \cong V\otimes U,\\
1\otimes U & \cong U\otimes1\cong U,\\
0\otimes U & \cong U\otimes0\cong0
\end{align*}
for any representations $U$, $V$ and $W$, and because tensor products preserve
isomorphisms (this means that if $U$, $V$ and $W$ are three representations of
$G$ such that $V\cong W$ (as representations), then $U\otimes V\cong U\otimes
W$ and $V\otimes U\cong W\otimes U$).
The ring $\operatorname*{FRep}_{k}G\diagup I$ is called the
\textit{representation ring of the group }$G$\textit{ over the field }$k$. The
elements of $\operatorname*{FRep}_{k}G\diagup I$ are called \textit{virtual
representations}.
This ring $\operatorname*{FRep}_{k}G\diagup I$ is helpful in working with
representations. However, its ring structure does not yet reflect everything
we can do with representations. In fact, we can build direct sums of
representations (this is addition in $\operatorname*{FRep}_{k}G\diagup I$) and
we can build tensor products (this is multiplication in $\operatorname*{FRep}%
_{k}G\diagup I$), but we can also build exterior powers of representations,
and we have no idea yet what operation on $\operatorname*{FRep}_{k}G\diagup I$
this entails. So we see that the abstract notion of a ring is not enough to
understand all of representation theory. We need a notion of a ring together
with some operations that ``behave like'' taking exterior powers. What axioms
should these operations satisfy?
Every representation $V$ of a group $G$ satisfies $\wedge^{0}V\cong1$ and
$\wedge^{1}V\cong V$. Besides, for any two representations $V$ and $W$ of $G$
and every $k\in\mathbb{N}$, there exists an isomorphism%
\begin{equation}
\wedge^{k}\left( V\oplus W\right) \cong\bigoplus_{i=0}^{k}\wedge^{i}%
V\otimes\wedge^{k-i}W \label{RepThV+W}%
\end{equation}
(see Exercise 1.1). In the representation ring, this means%
\[
\wedge^{k}\left( V+W\right) =\sum_{i=0}^{k}\left( \wedge^{i}V\right)
\cdot\left( \wedge^{k-i}W\right) .
\]
This already gives us three axioms for the operations that we want to
introduce. If we extend these three axioms to arbitrary elements of
$\operatorname*{FRep}_{k}G\diagup I$ (and not just actual representations), we
can compute $\wedge^{k}$ of virtual representations (and it turns out that it
is well-defined), and we obtain the notion of a $\lambda$\textit{-ring}.
We can still wonder whether these axioms are all that we can say about group
representations. The answer is no: In addition to the formula (\ref{RepThV+W}%
), there exist relations of the form
\begin{align}
& \wedge^{k}\left( V\otimes W\right) =P_{k}\left( \wedge^{1}V,\wedge
^{2}V,...,\wedge^{k}V,\wedge^{1}W,\wedge^{2}W,...,\wedge^{k}W\right)
\nonumber\\
& \ \ \ \ \ \ \ \ \ \ \text{for every }k\in\mathbb{N}\text{ and any two
representations }V\text{ and }W\text{ of }G \label{RepThVW}%
\end{align}
and%
\begin{align}
& \wedge^{k}\left( \wedge^{j}\left( V\right) \right) =P_{k,j}\left(
\wedge^{1}V,\wedge^{2}V,...,\wedge^{kj}V\right) \nonumber\\
& \ \ \ \ \ \ \ \ \ \ \text{for every }k\in\mathbb{N},\text{ }j\in
\mathbb{N}\text{ and any representation }V\text{ of }G, \label{RepThLL}%
\end{align}
where $P_{k}\in\mathbb{Z}\left[ \alpha_{1},\alpha_{2},...,\alpha_{k}%
,\beta_{1},\beta_{2},...,\beta_{k}\right] $ and $P_{k,j}\in\mathbb{Z}\left[
\alpha_{1},\alpha_{2},...,\alpha_{kj}\right] $ are ``universal'' polynomials
(i. e., polynomials only depending on $k$ resp. on $k$ and $j$, but not on
$V$, $W$ or $G$). These polynomials are rather hard to write down explicitly,
so it will need some theoretical preparation to define them.\footnote{Note
that these polynomials can have negative coefficients, so that the equality
(\ref{RepThVW}) does not necessarily mean an isomorphism of the kind%
\[
\wedge^{k}\left( V\otimes W\right) \cong\text{direct sum of some tensor
products of some }\wedge^{i}V\text{ and }\wedge^{j}W,
\]
but generally means an isomorphism of the kind%
\begin{align*}
& \wedge^{k}\left( V\otimes W\right) \oplus\text{direct sum of some tensor
products of some }\wedge^{i}V\text{ and }\wedge^{j}W\\
& \cong\text{(another) direct sum of some tensor products of some }\wedge
^{i}V\text{ and }\wedge^{j}W,
\end{align*}
and similarly (\ref{RepThLL}) has to be understood.}
These relations (\ref{RepThV+W}) and (\ref{RepThLL}), generalized to arbitrary
virtual representations, abstract to the notion of a \textit{special }%
$\lambda$\textit{-ring}. So$\ \operatorname*{FRep}_{k}G\diagup I$ is not just
a $\lambda$-ring; it is a special $\lambda$-ring. However, it has even more
structure than that: It is an \textit{augmented }$\lambda$\textit{-ring with
positive structure}. ``Augmented'' means the existence of a ring homomorphism
$\varepsilon:\operatorname*{FRep}_{k}G\diagup I\rightarrow\mathbb{Z}$ (a
so-called \textit{augmentation}) with certain properties; we will list these
properties later, but let us now notice that for our representation ring
$\operatorname*{FRep}_{k}G\diagup I$, the obvious natural choice of
$\varepsilon$ is the homomorphism which maps every representation $V$ of $G$
to $\dim V\in\mathbb{Z}$. A ``positive structure'' is a subset of $K$ closed
under addition and multiplication and containing $1$, and satisfying other
properties; in our case, the best choice for a positive structure on
$\operatorname*{FRep}_{k}G\diagup I$ is the subset%
\[
\left\{ \overline{V}\mid V\text{ is a representation of }G\right\}
\setminus0\subseteq\operatorname*{FRep}\nolimits_{k}G\diagup I.
\]
The reader may wonder how much the ring $\operatorname*{FRep}_{k}G\diagup I$
actually tells us about representations of $G$. For example, if $U$ and $V$
are two representations of $G$ such that $\overline{U}=\overline{V}$ in
$\operatorname*{FRep}_{k}G\diagup I$, does this mean that $U\cong V$ ? It
turns out that this is true, thanks to the cancellative property of
representation theory\footnote{This is the property that whenever $U$, $V$ and
$W$ are three representations of a finite group $G$ such that $U\oplus W\cong
V\oplus W$ (where we recall once again that ``representation'' means
``finite-dimensional representation'' for us!), then $U\cong V$. This can be
proven using the Krull-Remak-Schmidt theorem, or, when the characteristic of
the field is $0$, using semisimplicity of $k\left[ G\right] $.}; hence,
abstract algebraic identities that we can prove to hold in arbitrary special
$\lambda$-rings yield actual isomorphies of representations of finite groups.
(Of course, they only yield them once we will have proven that
$\operatorname*{FRep}_{k}G\diagup I$ is a special $\lambda$-ring. At the
moment, this is not proven in this text, although it is rather easy to show
using character theory.)
\subsection{Grothendieck rings of groups}
The situation gets more complicated when the field over which we are working
is not of characteristic $0$. In this case, it turns out that
$\operatorname*{FRep}_{k}G\diagup I$ is not necessarily a special $\lambda
$-ring any more (although still a $\lambda$-ring by Exercise 1.1). If we
insist on getting a special $\lambda$-ring, we must modify our definition of
$I$ to%
\begin{align*}
I & =\left\langle V-U-W\ \mid\ U\text{, }V\text{ and }W\text{ are three
representations of }G\text{ such that}\right. \\
& \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left. \text{there
exists an exact sequence }0\rightarrow U\rightarrow V\rightarrow
W\rightarrow0\right\rangle .
\end{align*}
The resulting ring $\operatorname*{FRep}_{k}G\diagup I$ is called the
\textit{Grothendieck ring} of representations of $G$ over our field. A proof
that it is a special $\lambda$-ring is sketched in \cite[Example on page
95]{Seiler88}, but I do not understand it. Anyway this result is not as strong
as in characteristic $0$ anymore, because the equality $\overline{U}%
=\overline{V}$ in the Grothendieck ring $\operatorname*{FRep}_{k}G\diagup I$
does not imply $U\cong V$ as representations of $G$ when $\operatorname*{char}%
k\neq0$. So the Grothendieck ring is, in some sense, a pale shadow of the
representation theory of $G$.
\subsection{Vector bundles}
Vector bundles over a given compact Hausdorff space are similar to
representations of a given group in several ways: They are somehow
``enriched'' vector space structures (a vector bundle is, roughly speaking, a
family of vector spaces with additional topological structure; a
representation of a group is a vector space with a group action on it), so one
can form direct sums, tensor products and exterior powers of both of these.
Hence, it is not surprising that we can define a $\lambda$-ring structure on a
kind of ``ring of vector bundles over a space'' similarly to the $\lambda
$-ring structure on the representation ring of a group. However, just as in
the case of representations of a group over nonzero characteristic, we must be
careful with vector bundles, because this ``ring of vector bundles over a
space'' actually does not consist of vector bundles, but of equivalence
classes, and sometimes, different vector bundles can lie in one and the same
equivalence class (just as representations of groups are no longer uniquely
determined by their equivalence class in the representation ring when the
characteristic of the ground field is not $0$). This ``ring of vector
bundles'' is denoted by $K\left( X\right) $, where $X$ is the base space,
and is the first fundamental object of study in K-theory. We will not delve
into K-theory here; we will only provide some of its backbone, namely the
abstract algebraic theory of $\lambda$-rings (which appear not only in
K-theory, but also in representation theory and elsewhere).
\subsection{Exercises}
\begin{quotation}
\textit{Exercise 1.1.} Let $G$ be a group, and let $V$ and $W$ be two
representations of $G$. Let $k\in\mathbb{N}$. Let $\iota_{V}:V\rightarrow
V\oplus W$ and $\iota_{W}:W\rightarrow V\oplus W$ be the canonical injections.
For every $i\in\left\{ 0,1,...,k\right\} $, we can define a vector space
homomorphism%
\[
\Phi_{i}:\wedge^{i}V\otimes\wedge^{k-i}W\rightarrow\wedge^{k}\left( V\oplus
W\right)
\]
by requiring that it sends
\begin{align*}
& \left( v_{1}\wedge v_{2}\wedge...\wedge v_{i}\right) \otimes\left(
w_{1}\wedge w_{2}\wedge...\wedge w_{k-i}\right) \ \ \ \ \ \ \ \ \ \ \text{to}%
\\
& \iota_{V}\left( v_{1}\right) \wedge\iota_{V}\left( v_{2}\right)
\wedge...\wedge\iota_{V}\left( v_{i}\right) \wedge\iota_{W}\left(
w_{1}\right) \wedge\iota_{W}\left( w_{2}\right) \wedge...\wedge\iota
_{W}\left( w_{k-i}\right)
\end{align*}
for all $v_{1},v_{2},...,v_{i}\in V$ and $w_{1},w_{2},...,w_{k-i}\in W$.
\textbf{(a)} Prove that this vector space homomorphism $\Phi_{i}$ is a
homomorphism of representations.
\textbf{(b)} Prove that the vector space homomorphism%
\[
\bigoplus_{i=0}^{k}\wedge^{i}V\otimes\wedge^{k-i}W\rightarrow\wedge^{k}\left(
V\oplus W\right)
\]
composed of the homomorphisms $\Phi_{i}$ for all $i\in\left\{
0,1,...,k\right\} $ is a canonical isomorphism of representations.
\end{quotation}
\section{$\lambda$-rings}
\subsection{The definition}
The following definition introduces our most important notions: that of a
$\lambda$-ring, that of a $\lambda$-ring homomorphism, and that of a
sub-$\lambda$-ring. While these notions are rather elementary (and much easier
to define than the ones in Sections 5 and later), they are the basis of our theory.
\begin{quote}
\textbf{Definition.} \textbf{1)} Let $K$ be a ring. Let $\lambda
^{i}:K\rightarrow K$ be a mapping\footnote{Here, \textquotedblleft
mapping\textquotedblright\ actually means \textquotedblleft
mapping\textquotedblright\ and not \textquotedblleft group
homomorphism\textquotedblright\ or \textquotedblleft ring
homomorphism\textquotedblright.} for every $i\in\mathbb{N}$ such that%
\begin{equation}
\lambda^{0}\left( x\right) =1\text{ and }\lambda^{1}\left( x\right)
=x\ \ \ \ \ \ \ \ \ \ \text{for every }x\in K. \label{lambda0}%
\end{equation}
Assume that%
\begin{equation}
\lambda^{k}\left( x+y\right) =\sum_{i=0}^{k}\lambda^{i}\left( x\right)
\lambda^{k-i}\left( y\right) \ \ \ \ \ \ \ \ \ \ \text{for every }%
k\in\mathbb{N},\text{ }x\in K\text{ and }y\in K. \label{lambda1}%
\end{equation}
Then, we call $\left( K,\left( \lambda^{i}\right) _{i\in\mathbb{N}}\right)
$ a $\lambda$\textit{-ring}. We will also call $K$ itself a $\lambda$-ring if
there is an obvious (from the context) choice of the sequence of mappings
$\left( \lambda^{i}\right) _{i\in\mathbb{N}}$ which makes $\left( K,\left(
\lambda^{i}\right) _{i\in\mathbb{N}}\right) $ a $\lambda$-ring.
\textbf{2)} Let $\left( K,\left( \lambda^{i}\right) _{i\in\mathbb{N}%
}\right) $ and $\left( L,\left( \mu^{i}\right) _{i\in\mathbb{N}}\right) $
be two $\lambda$-rings. Let $f:K\rightarrow L$ be a map. Then, $f$ is called a
$\lambda$\textit{-ring homomorphism} (or \textit{homomorphism of }$\lambda
$\textit{-rings}) if and only if $f$ is a ring homomorphism and satisfies
$\mu^{i}\circ f=f\circ\lambda^{i}$ for every $i\in\mathbb{N}$.
\textbf{3)} Let $\left( K,\left( \lambda^{i}\right) _{i\in\mathbb{N}%
}\right) $ be a $\lambda$-ring. Let $L$ be a subring of $K$. Then, $L$ is
said to be a \textit{sub-}$\lambda$\textit{-ring} of $\left( K,\left(
\lambda^{i}\right) _{i\in\mathbb{N}}\right) $ if and only if $\lambda
^{i}\left( L\right) \subseteq L$ for every $i\in\mathbb{N}$. Obviously, if
$L$ is a sub-$\lambda$-ring of $\left( K,\left( \lambda^{i}\right)
_{i\in\mathbb{N}}\right) $, then $\left( L,\left( \lambda^{i}\mid
_{L}\right) _{i\in\mathbb{N}}\right) $ is a $\lambda$-ring, and the
canonical inclusion $L\rightarrow K$ is a $\lambda$-ring homomorphism.
\end{quote}
\subsection{An alternative characterization}
We will now give an alternative characterization of $\lambda$-rings:
\begin{quote}
\textbf{Theorem 2.1.} Let $K$ be a ring. Let $\lambda^{i}:K\rightarrow K$ be a
mapping\footnote{Here, \textquotedblleft mapping\textquotedblright\ actually
means \textquotedblleft mapping\textquotedblright\ and not \textquotedblleft
group homomorphism\textquotedblright\ or \textquotedblleft ring
homomorphism\textquotedblright.} for every $i\in\mathbb{N}$ such that
$\lambda^{0}\left( x\right) =1$ and $\lambda^{1}\left( x\right) =x$ for
every $x\in K$. Consider the ring $K\left[ \left[ T\right] \right] $ of
formal power series in the indeterminate $T$ over the ring $K$. Define a map
$\lambda_{T}:K\rightarrow K\left[ \left[ T\right] \right] $ by
\[
\lambda_{T}\left( x\right) =\sum\limits_{i\in\mathbb{N}}\lambda^{i}\left(
x\right) T^{i}\ \ \ \ \ \ \ \ \ \ \text{for every }x\in K.
\]
Note that the power series $\lambda_{T}\left( x\right) =\sum\limits_{i\in
\mathbb{N}}\lambda^{i}\left( x\right) T^{i}$ has the coefficient
$\lambda^{0}\left( x\right) =1$ before $T^{0}$; thus, it is invertible in
$K\left[ \left[ T\right] \right] $.
\textbf{(a)} Then,
\[
\lambda_{T}\left( x\right) \cdot\lambda_{T}\left( y\right) =\lambda
_{T}\left( x+y\right) \ \ \ \ \ \ \ \ \ \ \text{for every }x\in K\text{ and
every }y\in K
\]
if and only if $\left( K,\left( \lambda^{i}\right) _{i\in\mathbb{N}%
}\right) $ is a $\lambda$-ring.
\textbf{(b)} Let $\left( K,\left( \lambda^{i}\right) _{i\in\mathbb{N}%
}\right) $ be a $\lambda$-ring. Then,%
\begin{align*}
\lambda_{T}\left( 0\right) & =1;\\
\lambda_{T}\left( -x\right) & =\left( \lambda_{T}\left( x\right)
\right) ^{-1}\ \ \ \ \ \ \ \ \ \ \text{for every }x\in K;\\
\lambda_{T}\left( x\right) \cdot\lambda_{T}\left( y\right) &
=\lambda_{T}\left( x+y\right) \ \ \ \ \ \ \ \ \ \ \text{for every }x\in
K\text{ and every }y\in K.
\end{align*}
\textbf{(c)} Let $\left( K,\left( \lambda^{i}\right) _{i\in\mathbb{N}%
}\right) $ and $\left( L,\left( \mu^{i}\right) _{i\in\mathbb{N}}\right) $
be two $\lambda$-rings. Consider the map $\lambda_{T}:K\rightarrow K\left[
\left[ T\right] \right] $ defined above, and a similarly defined map
$\mu_{T}:L\rightarrow L\left[ \left[ T\right] \right] $ for the $\lambda
$-ring $L$. Let $f:K\rightarrow L$ be a ring homomorphism. Consider the rings
$K\left[ \left[ T\right] \right] $ and $L\left[ \left[ T\right]
\right] $. Obviously, the homomorphism $f$ induces a homomorphism $f\left[
\left[ T\right] \right] :K\left[ \left[ T\right] \right] \rightarrow
L\left[ \left[ T\right] \right] $ (defined by
\begin{align*}
\left( f\left[ \left[ T\right] \right] \right) \left( \sum
\limits_{i\in\mathbb{N}}a_{i}T^{i}\right) & =\sum\limits_{i\in\mathbb{N}%
}f\left( a_{i}\right) T^{i}\\
& \ \ \ \ \ \ \ \ \ \ \text{for every }\sum\limits_{i\in\mathbb{N}}a_{i}%
T^{i}\in K\left[ \left[ T\right] \right] \text{ with }a_{i}\in K
\end{align*}
).
Then, $f$ is a $\lambda$-ring homomorphism if and only if $\mu_{T}\circ
f=f\left[ \left[ T\right] \right] \circ\lambda_{T}$.
\textbf{(d)} Let $\left( K,\left( \lambda^{i}\right) _{i\in\mathbb{N}%
}\right) $ be a $\lambda$-ring. Then, $\lambda^{i}\left( 0\right) =0$ for
every positive integer $i$.
\end{quote}
\begin{proof}
[Proof of Theorem 2.1.]\textbf{(a)} Every $x\in K$ and every $y\in K$ satisfy%
\begin{align*}
\lambda_{T}\left( x\right) \cdot\lambda_{T}\left( y\right) & =\left(
\sum\limits_{i\in\mathbb{N}}\lambda^{i}\left( x\right) T^{i}\right)
\cdot\left( \sum\limits_{i\in\mathbb{N}}\lambda^{i}\left( y\right)
T^{i}\right) \\
& \ \ \ \ \ \ \ \ \ \ \left( \text{since }\lambda_{T}\left( x\right)
=\sum\limits_{i\in\mathbb{N}}\lambda^{i}\left( x\right) T^{i}\text{ and
}\lambda_{T}\left( y\right) =\sum\limits_{i\in\mathbb{N}}\lambda^{i}\left(
y\right) T^{i}\right) \\
& =\sum_{k\in\mathbb{N}}\sum_{i=0}^{k}\lambda^{i}\left( x\right)
\lambda^{k-i}\left( y\right) \cdot T^{k}\\
& \ \ \ \ \ \ \ \ \ \ \left( \text{by the definition of the product of two
formal power series}\right)
\end{align*}
and%
\[
\lambda_{T}\left( x+y\right) =\sum_{i\in\mathbb{N}}\lambda^{i}\left(
x+y\right) T^{i}=\sum_{k\in\mathbb{N}}\lambda^{k}\left( x+y\right) T^{k}.
\]
Hence, the equation $\lambda_{T}\left( x\right) \cdot\lambda_{T}\left(
y\right) =\lambda_{T}\left( x+y\right) $ is equivalent to $\sum
\limits_{k\in\mathbb{N}}\sum\limits_{i=0}^{k}\lambda^{i}\left( x\right)
\lambda^{k-i}\left( y\right) \cdot T^{k}=\sum\limits_{k\in\mathbb{N}}%
\lambda^{k}\left( x+y\right) T^{k}$, which, in turn, means that every
$k\in\mathbb{N}$ satisfies $\sum\limits_{i=0}^{k}\lambda^{i}\left( x\right)
\lambda^{k-i}\left( y\right) =\lambda^{k}\left( x+y\right) $, and this is
exactly the property (\ref{lambda1}) from the definition of a $\lambda$-ring.
Thus, we have $\lambda_{T}\left( x\right) \cdot\lambda_{T}\left( y\right)
=\lambda_{T}\left( x+y\right) $ for every $x\in K$ and every $y\in K$ if and
only if $\left( K,\left( \lambda^{i}\right) _{i\in\mathbb{N}}\right) $ is
a $\lambda$-ring. This proves Theorem 2.1 \textbf{(a)}.
\textbf{(b)} Theorem 2.1 \textbf{(a)} tells us that $\lambda_{T}\left(
x\right) \cdot\lambda_{T}\left( y\right) =\lambda_{T}\left( x+y\right) $
for every $x\in K$ and every $y\in K$ if and only if $\left( K,\left(
\lambda^{i}\right) _{i\in\mathbb{N}}\right) $ is a $\lambda$-ring. Since we
know that $\left( K,\left( \lambda^{i}\right) _{i\in\mathbb{N}}\right) $
is a $\lambda$-ring, we thus conclude that
\begin{equation}
\lambda_{T}\left( x\right) \cdot\lambda_{T}\left( y\right) =\lambda
_{T}\left( x+y\right) \ \ \ \ \ \ \ \ \ \ \text{for every }x\in K\text{ and
every }y\in K. \label{2.1.pf.1}%
\end{equation}
Applied to $x=y=0$, this rewrites as $\lambda_{T}\left( 0\right)
\cdot\lambda_{T}\left( 0\right) =\lambda_{T}\left( 0+0\right) =\lambda
_{T}\left( 0\right) $, what yields $\lambda_{T}\left( 0\right) =1$ (since
$\lambda_{T}\left( 0\right) $ is invertible in $K\left[ \left[ T\right]
\right] $).
On the other hand, every $x\in K$ satisfies
\begin{align*}
\lambda_{T}\left( x\right) \cdot\lambda_{T}\left( -x\right) &
=\lambda_{T}\left( 0\right) \ \ \ \ \ \ \ \ \ \ \left( \text{by
(\ref{2.1.pf.1}), applied to }y=-x\right) \\
& =1,
\end{align*}
hence $\lambda_{T}\left( -x\right) =\left( \lambda_{T}\left( x\right)
\right) ^{-1}$. Theorem 2.1 \textbf{(b)} is thus proven.
\textbf{(c)} We have $\left( \mu_{T}\circ f\right) \left( x\right)
=\mu_{T}\left( f\left( x\right) \right) =\sum\limits_{i\in\mathbb{N}}%
\mu^{i}\left( f\left( x\right) \right) T^{i}$ (by the definition of
$\mu_{T}$) and $\left( f\left[ \left[ T\right] \right] \circ\lambda
_{T}\right) \left( x\right) =\left( f\left[ \left[ T\right] \right]
\right) \left( \lambda_{T}\left( x\right) \right) =\left( f\left[
\left[ T\right] \right] \right) \left( \sum\limits_{i\in\mathbb{N}%
}\lambda^{i}\left( x\right) T^{i}\right) =\sum\limits_{i\in\mathbb{N}%
}f\left( \lambda^{i}\left( x\right) \right) T^{i}$ for every $x\in K$.
Hence, $\mu_{T}\circ f=f\left[ \left[ T\right] \right] \circ\lambda_{T}$
is equivalent to $\sum\limits_{i\in\mathbb{N}}\mu^{i}\left( f\left(
x\right) \right) T^{i}=\sum\limits_{i\in\mathbb{N}}f\left( \lambda
^{i}\left( x\right) \right) T^{i}$ for every $x\in K$, which in turn is
equivalent to $\mu^{i}\left( f\left( x\right) \right) =f\left(
\lambda^{i}\left( x\right) \right) $ for every $x\in K$ and every
$i\in\mathbb{N}$, which in turn means that $\mu^{i}\circ f=f\circ\lambda^{i}$
for every $i\in\mathbb{N}$, which in turn means that $f$ is a $\lambda$-ring
homomorphism. This proves Theorem 2.1 \textbf{(c)}.
\textbf{(d)} Applying the equality $\lambda_{T}\left( x\right)
=\sum\limits_{i\in\mathbb{N}}\lambda^{i}\left( x\right) T^{i}$ to $x=0$, we
obtain $\lambda_{T}\left( 0\right) =\sum\limits_{i\in\mathbb{N}}\lambda
^{i}\left( 0\right) T^{i}$. But since $\lambda_{T}\left( 0\right) =1$,
this rewrites as $1=\sum\limits_{i\in\mathbb{N}}\lambda^{i}\left( 0\right)
T^{i}$. For every positive integer $i$, the coefficient of $T^{i}$ on the left
hand side of this equality is $0$, while the coefficient of $T^{i}$ on the
right hand side of this equality is $\lambda^{i}\left( 0\right) $. Since the
coefficients of $T^{i}$ on the two sides of an equality must be equal, this
yields $0=\lambda^{i}\left( 0\right) $ for every positive integer $i$. This
proves Theorem 2.1 \textbf{(d)}.
\end{proof}
The map $\lambda_{T}$ defined in Theorem 2.1 will follow us through the whole
theory of $\lambda$-rings. It is often easier to deal with than the maps
$\lambda^{i}$, since (as Theorem 2.1 \textbf{(a)} and \textbf{(b)} show)
$\lambda_{T}$ is a monoid homomorphism from $\left( K,+\right) $ to $\left(
K\left[ \left[ T\right] \right] ,\cdot\right) $ when $\left( K,\left(
\lambda^{i}\right) _{i\in\mathbb{N}}\right) $ is a $\lambda$-ring. Many
properties of $\lambda$-rings are easier to write in terms of $\lambda_{T}$
than in terms of the separate $\lambda^{i}$. We will later become acquainted
with the notion of ``special $\lambda$-rings'', for which $\lambda_{T}$ is not
only a monoid homomorphism but actually a $\lambda$-ring homomorphism (but not
to $K\left[ \left[ T\right] \right] $ but to a different $\lambda$-ring
with a new ring structure).
\subsection{$\lambda$-ideals}
Just as rings have ideals and Lie algebras have Lie ideals, there is a notion
of $\lambda$-ideals defined for $\lambda$-rings. Here is one way to define them:
\begin{quote}
\textbf{Definition.} Let $\left( K,\left( \lambda^{i}\right) _{i\in
\mathbb{N}}\right) $ be a $\lambda$-ring. Let $I$ be an ideal of the ring
$K$. Then, $I$ is said to be a $\lambda$\textit{-ideal} of $K$ if and only if
every $t\in I$ and every positive integer $i$ satisfy $\lambda^{i}\left(
t\right) \in I$.
\end{quote}
Just as rings can be factored by ideals to obtain new rings, and Lie algebras
can be factored by Lie ideals to obtain new Lie algebras, we can factor
$\lambda$-rings by $\lambda$-ideals and obtain new $\lambda$-rings:
\begin{quote}
\textbf{Theorem 2.2.} Let $\left( K,\left( \lambda^{i}\right)
_{i\in\mathbb{N}}\right) $ be a $\lambda$-ring. Let $I$ be a $\lambda$-ideal
of the ring $K$. For every $z\in K$, let $\overline{z}$ denote the residue
class of $z$ modulo $I$. (This $\overline{z}$ lies in $K\diagup I$.)
\textbf{(a)} If $x\in K\diagup I$ is arbitrary, and $y\in K$ and $z\in K$ are
two elements of $K$ satisfying $\overline{y}=x$ and $\overline{z}=x$, then
$\overline{\lambda^{i}\left( y\right) }=\overline{\lambda^{i}\left(
z\right) }$ for every $i\in\mathbb{N}$.
\textbf{(b)} For every $i\in\mathbb{N}$, define a map $\widetilde{\lambda}%
^{i}:K\diagup I\rightarrow K\diagup I$ as follows: For every $x\in K\diagup
I$, let $\widetilde{\lambda}^{i}\left( x\right) $ be defined as
$\overline{\lambda^{i}\left( w\right) }$, where $w$ is an element of $K$
satisfying $\overline{w}=x$. (This is well-defined because the value of
$\overline{\lambda^{i}\left( w\right) }$ does not depend on the choice of
$w$\ \ \ \ \footnote{In fact, any two choices of $w$ lead to the same value of
$\overline{\lambda^{i}\left( w\right) }$ (this follows from Theorem 2.2
\textbf{(a)}).}.)
Then, $\left( K\diagup I,\left( \widetilde{\lambda}^{i}\right)
_{i\in\mathbb{N}}\right) $ is a $\lambda$-ring.
\textbf{(c)} The canonical projection $K\rightarrow K\diagup I$ is a $\lambda
$-ring homomorphism.
\end{quote}
The proof of Theorem 2.2 is given in the solution of Exercise 2.3.
Along with Theorem 2.2 comes the following result:
\begin{quote}
\textbf{Theorem 2.3.} Let $\left( K,\left( \lambda^{i}\right)
_{i\in\mathbb{N}}\right) $ and $\left( L,\left( \mu^{i}\right)
_{i\in\mathbb{N}}\right) $ be two $\lambda$-rings. Let $f:K\rightarrow L$ be
a $\lambda$-ring homomorphism. Then, $\operatorname*{Ker}f$ is a $\lambda$-ideal.
\end{quote}
The proof of Theorem 2.3 is given in the solution of Exercise 2.4.
\subsection{Exercises}
\begin{quotation}
\textit{Exercise 2.1.} Let $\left( K,\left( \lambda^{i}\right)
_{i\in\mathbb{N}}\right) $ and $\left( L,\left( \mu^{i}\right)
_{i\in\mathbb{N}}\right) $ be two $\lambda$-rings. Let $f:K\rightarrow L$ be
a ring homomorphism. Let $E$ be a generating set of the $\mathbb{Z}$-module
$K$.
\textbf{(a)} Prove that $f$ is a $\lambda$-ring homomorphism if and only if
every $e\in E$ satisfies $\left( \mu_{T}\circ f\right) \left( e\right)
=\left( f\left[ \left[ T\right] \right] \circ\lambda_{T}\right) \left(
e\right) $.
\textbf{(b)} Prove that $f$ is a $\lambda$-ring homomorphism if and only if
every $e\in E$ satisfies $\left( \mu^{i}\circ f\right) \left( e\right)
=\left( f\circ\lambda^{i}\right) \left( e\right) $ for every
$i\in\mathbb{N}$.
\textit{Exercise 2.2.} Let $\left( K,\left( \lambda^{i}\right)
_{i\in\mathbb{N}}\right) $ be a $\lambda$-ring. Let $L$ be a subset of $K$
which is closed under addition, multiplication and the maps $\lambda^{i}$.
Assume that $0\in L$ and $1\in L$. Then, the subset $L-L$ of $K$ (this subset
$L-L$ is defined by $L-L=\left\{ \ell-\ell^{\prime}\mid\ell\in L,\ \ell
^{\prime}\in L\right\} $) is a sub-$\lambda$-ring of $K$.
\textit{Exercise 2.3.} Prove Theorem 2.2.
\textit{Exercise 2.4.} Prove Theorem 2.3.
\end{quotation}
\section{Examples of $\lambda$-rings}
\subsection{Binomial $\lambda$-rings}
Before we go deeper into the theory, it is time for some examples.
Obviously, the trivial ring $0$ (the ring satisfying $0=1$) along with the
trivial maps $\lambda^{i}:0\rightarrow0$ is a $\lambda$-ring. Let us move on
to more surprising examples:
\begin{quote}
\textbf{Theorem 3.1.} For every $i\in\mathbb{N}$, define a map $\lambda
^{i}:\mathbb{Z}\rightarrow\mathbb{Z}$ by $\lambda^{i}\left( x\right)
=\dbinom{x}{i}$ for every $x\in\mathbb{Z}$.\ \ \ \ \footnote{Note that
$\dbinom{x}{i}$ is defined to be $\dfrac{x\cdot\left( x-1\right)
\cdot...\cdot\left( x-i+1\right) }{i!}$ for every $x\in\mathbb{R}$ and
$i\in\mathbb{N}$.} Then, $\left( \mathbb{Z},\left( \lambda^{i}\right)
_{i\in\mathbb{N}}\right) $ is a $\lambda$-ring.
\end{quote}
\begin{proof}
[Proof of Theorem 3.1.]Trivially, $\lambda^{0}\left( x\right) =1$ and
$\lambda^{1}\left( x\right) =x$ for every $x\in\mathbb{Z}$. The only
challenge, if there is a challenge in this proof, is to verify the identity
(\ref{lambda1}) for $K=\mathbb{Z}$. In other words, we have to prove that%
\begin{equation}
\dbinom{x+y}{k}=\sum_{i=0}^{k}\dbinom{x}{i}\dbinom{y}{k-i} \label{vandermonde}%
\end{equation}
for every $k\in\mathbb{N}$, $x\in\mathbb{Z}$ and $y\in\mathbb{Z}$. This is the
so-called \textit{Vandermonde convolution identity}, and various proofs of it
can easily be found in the literature\footnote{For example, it follows
immediately from \cite[Theorem 3.29]{Grin-detn}.}. Probably the shortest proof
of (\ref{vandermonde}) is the following: If we fix $k\in\mathbb{N}$, then
(\ref{vandermonde}) is a polynomial identity in both $x$ and $y$ (indeed, both
sides of (\ref{vandermonde}) are polynomials in $x$ and $y$ with rational
coefficients), and thus it is enough to prove it for all natural $x$ and $y$
(because a polynomial identity holding for all natural variables must hold
everywhere). But for $x$ and $y$ natural, we have%
\begin{align*}
\sum_{k=0}^{x+y}\sum_{i=0}^{k}\dbinom{x}{i}\dbinom{y}{k-i}T^{k} &
=\underbrace{\sum_{i=0}^{x}\dbinom{x}{i}T^{i}}_{\substack{=\left( 1+T\right)
^{x}\\\text{(by the}\\\text{binomial formula)}}}\cdot\underbrace{\sum
_{j=0}^{y}\dbinom{y}{j}T^{j}}_{\substack{=\left( 1+T\right) ^{y}\\\text{(by
the}\\\text{binomial formula)}}}\\
& =\left( 1+T\right) ^{x}\cdot\left( 1+T\right) ^{y}=\left( 1+T\right)
^{x+y}=\sum_{k=0}^{x+y}\dbinom{x+y}{k}T^{k}\\
& \ \ \ \ \ \ \ \ \ \ \left( \text{by the binomial formula}\right)
\end{align*}
in the polynomial ring $\mathbb{Z}\left[ T\right] $. Comparing coefficients
before $T^{k}$ in this equality, we quickly conclude that (\ref{vandermonde})
holds for each $k\in\mathbb{N}$. Thus, (\ref{vandermonde}) is
proven.\footnote{\textit{Remark.} It is tempting to apply this argument to the
general case (where $x$ and $y$ are not required to be natural), because the
binomial formula holds for negative exponents as well (of course, this
requires working in the formal power series ring $\mathbb{Z}\left[ \left[
T\right] \right] $ rather than in the polynomial ring $\mathbb{Z}\left[
T\right] $), but I am not sure whether this argument is free of circular
reasoning because it is not at all obvious that $\left( 1+T\right)
^{x}\left( 1+T\right) ^{y}=\left( 1+T\right) ^{x+y}$ in $\mathbb{Z}\left[
\left[ T\right] \right] $ for negative $x$ and $y$, and I even fear that
this is usually proven using (\ref{vandermonde}).} This proves Theorem 3.1.
\end{proof}
Our next example is a generalization of Theorem 3.1:
\begin{quote}
\textbf{Definition.} Let $K$ be a ring. We call $K$ a \textit{binomial ring}
if and only if none of the elements $1$, $2$, $3$, $...$ is a zero-divisor in
$K$, and $n!\mid x\cdot\left( x-1\right) \cdot...\cdot\left( x-i+1\right)
$ for every $x\in K$ and every $n\in\mathbb{N}$.
\textbf{Theorem 3.2.} Let $K$ be a binomial ring. For every $i\in\mathbb{N}$,
define a map $\lambda^{i}:K\rightarrow K$ by $\lambda^{i}\left( x\right)
=\dbinom{x}{i}$ for every $x\in K$ (where, again, $\dbinom{x}{i}$ is defined
to be $\dfrac{x\cdot\left( x-1\right) \cdot...\cdot\left( x-i+1\right)
}{i!}$). Then, $\left( K,\left( \lambda^{i}\right) _{i\in\mathbb{N}%
}\right) $ is a $\lambda$-ring.
\end{quote}
Such $\lambda$-rings $K$ are called \textit{binomial }$\lambda$\textit{-rings}.
\begin{proof}
[Proof of Theorem 3.2.]Obviously, $\lambda^{0}\left( x\right) =1$ and
$\lambda^{1}\left( x\right) =x$ for every $x\in K$, so it only remains to
show that (\ref{lambda1}) is satisfied. This means proving (\ref{vandermonde})
for every $k\in\mathbb{N}$, $x\in K$ and $y\in K$. But this is easy now: Fix
$k\in\mathbb{N}$. Then, (\ref{vandermonde}) is a polynomial identity in both
$x$ and $y$, and since we know that it holds for every $x\in\mathbb{Z}$ and
every $y\in\mathbb{Z}$ (as we have seen in the proof of Theorem 3.1), it
follows that it holds for every $x\in K$ and every $y\in K$ (since a
polynomial identity holding for all integer variables must hold everywhere).
This completes the proof of Theorem 3.2.
\end{proof}
Obviously, the $\lambda$-ring $\left( \mathbb{Z},\left( \lambda^{i}\right)
_{i\in\mathbb{N}}\right) $ defined in Theorem 3.1 is a binomial $\lambda
$-ring. For other examples of binomial $\lambda$-rings, see Exercise 3.1. Of
course, every $\mathbb{Q}$-algebra is a binomial ring as well.
\subsection{Adjoining a polynomial variable to a $\lambda$-ring}
Binomial $\lambda$-rings are not the main examples of $\lambda$-rings. We will
see an important example of $\lambda$-rings in Theorem 5.1 and Exercise 6.1.
Another simple way to construct new examples from known ones is the following one:
\begin{quote}
\textbf{Definition.} Let $K$ be a ring. Let $L$ be a $K$-algebra. Consider the
ring $K\left[ \left[ T\right] \right] $ of formal power series in the
indeterminate $T$ over the ring $K$, and the ring $L\left[ \left[ T\right]
\right] $ of formal power series in the indeterminate $T$ over the ring $L$.
For every $\mu\in L$, we can define a $K$-algebra homomorphism
$\operatorname*{ev}_{\mu T}:K\left[ \left[ T\right] \right] \rightarrow
L\left[ \left[ T\right] \right] $ by setting $\operatorname*{ev}_{\mu
T}\left( \sum\limits_{i\in\mathbb{N}}a_{i}T^{i}\right) =\sum\limits_{i\in
\mathbb{N}}a_{i}\mu^{i}T^{i}$ for every power series $\sum\limits_{i\in
\mathbb{N}}a_{i}T^{i}\in K\left[ \left[ T\right] \right] $ (which
satisfies $a_{i}\in K$ for every $i\in\mathbb{N}$). (In other words,
$\operatorname*{ev}_{\mu T}$ is the map that takes any power series in $T$ and
replaces every $T$ in this power series by $\mu T$.)
\textbf{Theorem 3.3.} Let $\left( K,\left( \lambda^{i}\right)
_{i\in\mathbb{N}}\right) $ be a $\lambda$-ring. Consider the polynomial ring
$K\left[ S\right] $. For every $i\in\mathbb{N}$, define a map $\overline
{\lambda}^{i}:K\left[ S\right] \rightarrow K\left[ S\right] $ as follows:
For every $\sum\limits_{j\in\mathbb{N}}a_{j}S^{j}\in K\left[ S\right] $
(with $a_{j}\in K$ for every $j\in\mathbb{N}$), let $\overline{\lambda}%
^{i}\left( \sum\limits_{j\in\mathbb{N}}a_{j}S^{j}\right) $ be the
coefficient of the power series $\prod\limits_{j\in\mathbb{N}}\lambda_{S^{j}%
T}\left( a_{j}\right) \in\left( K\left[ S\right] \right) \left[ \left[
T\right] \right] $ before $T^{i}$, where the power series $\lambda_{S^{j}%
T}\left( a_{j}\right) \in\left( K\left[ S\right] \right) \left[ \left[
T\right] \right] $ is defined as $\operatorname*{ev}_{S^{j}T}\left(
\lambda_{T}\left( a_{j}\right) \right) $.
\textbf{(a)} Then, $\left( K\left[ S\right] ,\left( \overline{\lambda}%
^{i}\right) _{i\in\mathbb{N}}\right) $ is a $\lambda$-ring. The ring $K$ is
a sub-$\lambda$-ring of $\left( K\left[ S\right] ,\left( \overline
{\lambda}^{i}\right) _{i\in\mathbb{N}}\right) $.
\textbf{(b)} For every $a\in K$ and $\alpha\in\mathbb{N}$, we have
$\overline{\lambda}^{i}\left( aS^{\alpha}\right) =\lambda^{i}\left(
a\right) S^{\alpha i}$ for every $i\in\mathbb{N}$.
\end{quote}
\begin{proof}
[Proof of Theorem 3.3.]For every $x\in K$, we have
\begin{align}
\lambda_{S^{j}T}\left( x\right) & =\sum\limits_{i\in\mathbb{N}}\lambda
^{i}\left( x\right) \left( S^{j}T\right) ^{i}\ \ \ \ \ \ \ \ \ \ \left(
\text{since }\lambda_{T}\left( x\right) =\sum\limits_{i\in\mathbb{N}}%
\lambda^{i}\left( x\right) T^{i}\right) \nonumber\\
& =\underbrace{\lambda^{0}\left( x\right) }_{\substack{=1\\\text{(by
(\ref{lambda0}))}}}\underbrace{\left( S^{j}T\right) ^{0}}_{=1}%
+\underbrace{\lambda^{1}\left( x\right) }_{\substack{=x\\\text{(by
(\ref{lambda0}))}}}\underbrace{\left( S^{j}T\right) ^{1}}_{=S^{j}T}%
+\sum\limits_{i\geq2}\lambda^{i}\left( x\right) \underbrace{\left(
S^{j}T\right) ^{i}}_{=S^{ji}T^{i}}\nonumber\\
& =1+xS^{j}T+\underbrace{\sum\limits_{i\geq2}\lambda^{i}\left( x\right)
S^{ji}T^{i}}_{\substack{=\left( \text{sum of terms divisible by }%
T^{2}\right) \\\text{(since }T^{2}\mid T^{i}\text{ for every }i\geq2\text{)}%
}}\nonumber\\
& =1+xS^{j}T+\left( \text{sum of terms divisible by }T^{2}\right) .
\label{thm.3.3.pf.1}%
\end{align}
\textbf{(a)} Define a map $\overline{\lambda}_{T}:K\left[ S\right]
\rightarrow\left( K\left[ S\right] \right) \left[ \left[ T\right]
\right] $ by%
\begin{equation}
\overline{\lambda}_{T}\left( u\right) =\sum\limits_{i\in\mathbb{N}}%
\overline{\lambda}^{i}\left( u\right) T^{i}\ \ \ \ \ \ \ \ \ \ \text{for
every }u\in K\left[ S\right] . \label{thm.3.3.pf.a.1}%
\end{equation}
Then, according to the definition of the maps $\overline{\lambda}^{i}$, we
have
\begin{equation}
\overline{\lambda}_{T}\left( \sum\limits_{j\in\mathbb{N}}a_{j}S^{j}\right)
=\prod\limits_{j\in\mathbb{N}}\lambda_{S^{j}T}\left( a_{j}\right) \in\left(
K\left[ S\right] \right) \left[ \left[ T\right] \right]
\label{thm.3.3.pf.a.lambdaol}%
\end{equation}
for every $\sum\limits_{j\in\mathbb{N}}a_{j}S^{j}\in K\left[ S\right] $
(with $a_{j}\in K$ for every $j\in\mathbb{N}$). Hence, for every $u\in
K\left[ S\right] $, we have%
\begin{equation}
\sum\limits_{i\in\mathbb{N}}\overline{\lambda}^{i}\left( u\right)
T^{i}=1+uT+\left( \text{sum of terms divisible by }T^{2}\right)
\label{thm.3.3.pf.a.2}%
\end{equation}
in $\left( K\left[ S\right] \right) \left[ \left[ T\right] \right]
$\ \ \ \ \footnote{\textit{Proof of (\ref{thm.3.3.pf.a.2}):} Let $u\in
K\left[ S\right] $. Write $u$ in the form $u=\sum\limits_{j\in\mathbb{N}%
}a_{j}S^{j}$, where $a_{j}\in K$ for every $j\in\mathbb{N}$. Then,
(\ref{thm.3.3.pf.a.1}) yields%
\begin{align*}
\sum\limits_{i\in\mathbb{N}}\overline{\lambda}^{i}\left( u\right) T^{i} &
=\overline{\lambda}_{T}\left( \underbrace{u}_{=\sum\limits_{j\in\mathbb{N}%
}a_{j}S^{j}}\right) =\overline{\lambda}_{T}\left( \sum\limits_{j\in
\mathbb{N}}a_{j}S^{j}\right) =\prod\limits_{j\in\mathbb{N}}%
\underbrace{\lambda_{S^{j}T}\left( a_{j}\right) }_{\substack{=1+a_{j}%
S^{j}T+\left( \text{sum of terms divisible by }T^{2}\right) \\\text{(by
(\ref{thm.3.3.pf.1}), applied to }x=a_{j}\text{)}}}\\
& =\prod\limits_{j\in\mathbb{N}}\left( 1+a_{j}S^{j}T+\left( \text{sum of
terms divisible by }T^{2}\right) \right) \\
& =1+\underbrace{\left( \sum\limits_{j\in\mathbb{N}}a_{j}S^{j}\right)
}_{=u}T+\left( \text{sum of terms divisible by }T^{2}\right) \\
& =1+uT+\left( \text{sum of terms divisible by }T^{2}\right) .
\end{align*}
This proves (\ref{thm.3.3.pf.a.2}).}. Hence, for every $u\in K\left[
S\right] $, we have $\overline{\lambda}^{0}\left( u\right) =1$ (this is
obtained by comparing coefficients before $T^{0}$ in the equality
(\ref{thm.3.3.pf.a.2})) and $\overline{\lambda}^{1}\left( u\right) =u$ (this
is obtained by comparing coefficients before $T^{1}$ in the equality
(\ref{thm.3.3.pf.a.2})). Renaming $u$ as $x$ in this sentence, we obtain the
following: For every $x\in K\left[ S\right] $, we have $\overline{\lambda
}^{0}\left( x\right) =1$ and $\overline{\lambda}^{1}\left( x\right) =x$.
Thus, we can apply Theorem 2.1 \textbf{(a)} to $K\left[ S\right] $, $\left(
\overline{\lambda}^{i}\right) _{i\in\mathbb{N}}$ and $\overline{\lambda}_{T}$
instead of $K$, $\lambda^{i}$ and $\lambda_{T}$. As a result, we see that
\begin{equation}
\overline{\lambda}_{T}\left( x\right) \cdot\overline{\lambda}_{T}\left(
y\right) =\overline{\lambda}_{T}\left( x+y\right)
\ \ \ \ \ \ \ \ \ \ \text{for every }x\in K\left[ S\right] \text{ and every