Characters of Modules over Finite-Dimensional Algebras

Q: Why does the character only see composition factors?

Because trace is additive along block-triangular decompositions. Choosing a basis adapted to a submodule makes every action block upper triangular, and the trace is the sum of the diagonal blocks' traces. Extension data lives in the off-diagonal block, which the trace ignores.

Q: Is a character determined by its values on a basis of R?

Yes, since it is k-linear. For a group algebra it is determined by its values on group elements, and in fact only by the values on conjugacy class representatives, since it vanishes on all commutators.

Q: Why is (7.20) true in characteristic p when (7.19) is not?

Because it needs only that m_1 is nonzero in k, which forces m_1 ≥ 1 as an integer. (7.19) needs to divide by _k M_i, which may be zero in k. The absolute irreducibility of M is what lets one prescribe the trace to be 1 rather than _k M_1.

Q: What replaces (7.19) in characteristic p?

Brauer character theory. One evaluates on p-regular elements, lifts the eigenvalues of the action to roots of unity in characteristic 0, and sums them there. The resulting functions do determine composition factors of modules in characteristic p.

Q: Do characters of non-isomorphic simple modules have to differ?

Over a splitting field yes, by (7.21), in any characteristic. Over a general field they can coincide only if the modules fail to be absolutely irreducible, so the first place to look for pathologies is a field over which the algebra does not split.

Q: How many independent characters can an algebra have?

At most _k R/[R,R], since every character is a trace function. When R splits the simple characters already achieve _k R/T(R) = r, and they form a basis of the functionals annihilating T(R).

Executive Summary

A module $M$ of finite $k$ -dimension over a finite-dimensional $k$ -algebra $R$ has a character $χ_{M} : R \to k$ , $χ_{M} (a) = tr (a acting on M)$ . It is a single linear functional, and it compresses an entire module into $\dim_{k} R$ scalars.

Characters are additive on short exact sequences, so they see only composition factors. In characteristic $0$ that is all they lose: $χ_{M}$ recovers the multiplicities exactly $(7.19)$ . In characteristic $p$ even that fails, because multiplicities are only determined modulo $p$ — but absolutely irreducible modules are still separated $(7.20)$ , which is what modular representation theory needs.

$χ_{M} (a) = tr (a ∣ M)$ Definition

additiveOn short exact sequences

$char k = 0$ Composition factors recovered

$r$ Independent characters when

R

splits

Overview

Let $R$ be a $k$ -algebra with $\dim_{k} R < \infty$ and let $M$ be a left $R$ -module with $\dim_{k} M < \infty$ . Each $a \in R$ acts on $M$ as a $k$ -linear map, and its trace is a scalar; the assignment $a \mapsto χ_{M} (a)$ is $k$ -linear, satisfies $χ_{M} (a b) = χ_{M} (b a)$ , and takes the value $\dim_{k} M \cdot 1_{k}$ at $a = 1$ .

The proofs on this page all use the same device: because $R / rad R$ is a product of simple components, one can select an element of $R$ that acts as a chosen endomorphism on one simple module and as zero on the others. Evaluating a character at such an element isolates one multiplicity. Everything else is arithmetic in $k$ .

For a group algebra $k G$ the restriction of $χ_{M}$ to $G$ is the classical character of the representation, and the theorems here specialise to statements every representation theorist uses daily. The orthogonality relations, treated on Orthogonality Relations, are the quantitative refinement available in the semisimple case.

Learning Objectives

Define $χ_{M}$ and verify linearity, the trace identity and the value at $1$ .
Prove additivity along $0 \to M' \to M \to M'' \to 0$ using a block-triangular basis.
Prove $(7.19)$ and locate exactly where $char k = 0$ is used.
Give a characteristic zero example of non-isomorphic modules with equal characters.
Prove $(7.20)$ and deduce that characters separate simple modules over a splitting field.
Show that the characters of the simple modules form a basis of the functionals killing $T (R)$ .

Definitions

DefinitionCharacter of a module

Let $R$ be a $k$ -algebra with $\dim_{k} R < \infty$ and $M$ a left $R$ -module with $\dim_{k} M < \infty$ . For $a \in R$ let $λ_{a} : M \to M$ be the $k$ -linear map $m \mapsto a m$ . The character of $M$ is

χ_{M} : R ⟶ k, χ_{M} (a) = tr (λ_{a})

(7.19a)

It is $k$ -linear; it satisfies $χ_{M} (a b) = χ_{M} (b a)$ , hence vanishes on $[R, R]$ ; and $χ_{M} (1) = (\dim_{k} M) \cdot 1_{k}$ . It also vanishes on $rad R$ whenever $M$ is semisimple, since then the radical acts as zero.

$χ_{M}$: The character of $M$ ; for $R = k G$ its restriction to $G$ is the classical character of the corresponding representation.
$m_{i}$: The multiplicity of the simple module $M_{i}$ as a composition factor of $M$ , a non-negative integer.
$a_{i}$: An element of $R$ whose image in $R / rad R$ is the identity of the $i$ -th simple component: it acts as the identity on $M_{i}$ and as zero on $M_{j}$ for $j \neq i$ .
$T (R)$: $rad R + [R, R]$ ; every character of a semisimple module vanishes on it.

Characters are defined for arbitrary finite-dimensional modules, not only semisimple ones. It is the conclusions, not the definition, that need semisimplicity.

Core Concepts

Additivity, and what it costs

Let $0 \to M' \to M \to M'' \to 0$ be exact with all three modules of finite $k$ -dimension. Choosing a basis of $M$ extending one of $M'$ makes every $λ_{a}$ block upper triangular with diagonal blocks the actions on $M'$ and $M''$ , so the traces add:

χ_{M} = χ_{M'} + χ_{M''}, hence χ_{M} = \sum_{i = 1}^{r} m_{i} χ_{M_{i}}

(7.19b)

$m_{i}$ is the multiplicity of the simple module $M_{i}$ as a composition factor of $M$ .

So the character factors through the Grothendieck group: it is a function of the composition factors alone. This is a genuine loss of information — extensions are invisible — and is the reason $(7.19)$ concludes with isomorphism only for semisimple modules.

Isolating a multiplicity

Because $R ↠ R / rad R ≅ B_{1} \times \dots \times B_{r}$ , the components can be addressed independently. Evaluating $(7.19 b)$ at $a_{i}$ gives $χ_{M} (a_{i}) = m_{i} (\dim_{k} M_{i})$ in $k$ : one equation, one unknown. Whether it can be solved for $m_{i}$ is exactly a question about the characteristic.

Key Results

Theorem(7.19)Characters in characteristic zero

Let $R$ be a $k$ -algebra with $\dim_{k} R < \infty$ and $char k = 0$ , and let $M$ be a left $R$ -module with $\dim_{k} M < \infty$ . Then $χ_{M}$ determines the composition factors of $M$ together with their multiplicities.

Consequently, if $M, M'$ are left $R$ -modules of finite $k$ -dimension with $χ_{M} = χ_{M'}$ , then $M$ and $M'$ have the same composition factors with the same multiplicities; and if both are semisimple, then $M ≅ M'$ .

Proof

In the notation of $(7.1)$ , let $m_{i} \geq 0$ be the multiplicity of $M_{i}$ among the composition factors of $M$ , so that $χ_{M} = \sum_{i} m_{i} χ_{M_{i}}$ by additivity.

Choose $a_{i} \in R$ whose image in $R / rad R$ is the identity element of the $i$ -th simple component. Then $a_{i}$ acts as zero on $M_{j}$ for $j \neq i$ , so $χ_{M_{j}} (a_{i}) = 0$ ; and it acts as the identity on $M_{i}$ , so $χ_{M_{i}} (a_{i}) = (\dim_{k} M_{i}) \cdot 1_{k}$ . Evaluating,

χ_{M} (a_{i}) = m_{i} (\dim_{k} M_{i}) in k

Since $char k = 0$ , the prime field is $ℚ$ and the nonzero integer $\dim_{k} M_{i}$ is invertible in $k$ ; hence $m_{i} = χ_{M} (a_{i}) / \dim_{k} M_{i}$ , and this identity determines the integer $m_{i}$ because distinct non-negative integers remain distinct in $k$ .

So all multiplicities are functions of $χ_{M}$ . If $χ_{M} = χ_{M'}$ the two modules have identical multiplicities, hence identical composition factors. If moreover both are semisimple, each is the direct sum of its composition factors with those multiplicities, so $M ≅ M'$ .

CounterexampleSemisimplicity cannot be dropped

Let $k$ have characteristic $0$ and $R = k [t] / (t^{2})$ , of dimension $2$ . Let $M = k e_{1} \oplus k e_{2}$ with $t e_{1} = e_{2}$ , $t e_{2} = 0$ — this is the regular module — and let $M' = k^{2}$ with $t$ acting as zero.

On $M$ , the element $t$ acts nilpotently, so $χ_{M} (a + b t) = 2 a$ ; the same computation on $M'$ gives $χ_{M'} (a + b t) = 2 a$ . The characters agree, and indeed both modules have two composition factors, each the unique simple module $k$ . But $M ≅_{R} R$ is indecomposable, since $R$ is local and its only idempotents are $0$ and $1$ , whereas $M'$ is semisimple. So the two modules are not isomorphic.

CounterexampleCharacteristic

p

: even the dimension is lost

Let $char k = p > 0$ and let $M_{1}, M_{2}$ be non-isomorphic simple $R$ -modules. Then $M = M_{1}^{\oplus p}$ and $M' = M_{2}^{\oplus p}$ both have character $p χ_{M_{i}} = 0$ . They have no composition factor in common, and their $k$ -dimensions may differ.

A concrete instance: for $R = k C_{p}$ with $char k = p$ , the regular module has character $χ (g) = 0$ for $g \neq 1$ and $χ (1) = p \cdot 1_{k} = 0$ , so its character is identically zero — the same as that of the zero module.

Theorem(7.20)Characters separate absolutely irreducible modules

Let $R$ be a $k$ -algebra with $\dim_{k} R < \infty$ , and let $M, M'$ be left $R$ -modules of finite $k$ -dimension with $M$ absolutely irreducible. Assume either

$\dim_{k} M = \dim_{k} M'$ , or
$M'$ is irreducible.

Then $M ≅ M'$ if and only if $χ_{M} = χ_{M'}$ . The characteristic of $k$ is arbitrary.

Proof

Isomorphic modules obviously have the same character, so only the converse needs proof. Assume $χ_{M} = χ_{M'}$ and, in the notation of $(7.1)$ , that $M = M_{1}$ . Let $m_{i}$ be the multiplicity of $M_{i}$ among the composition factors of $M'$ .

Since $M_{1}$ is absolutely irreducible, the map $R \to B_{1} = End (M_{1})_{k}$ is surjective by $(7.5) (2)$ . As $R ↠ \prod_{i} B_{i}$ , we may choose $a \in R$ projecting to an endomorphism of $M_{1}$ of trace $1$ — for instance a rank-one idempotent — and projecting to $0$ in $B_{i}$ for every $i \geq 2$ .

Then $χ_{M_{1}} (a) = 1$ and $χ_{M_{i}} (a) = 0$ for $i \geq 2$ , so

1 = χ_{M} (a) = χ_{M'} (a) = \sum_{i = 1}^{r} m_{i} χ_{M_{i}} (a) = m_{1} in k

In particular $m_{1} \neq 0$ in $k$ , so $m_{1} \geq 1$ as an integer: $M_{1}$ really occurs among the composition factors of $M'$ .

Under hypothesis (2), $M'$ is simple and has $M_{1}$ as a composition factor, so $M' ≅ M_{1} = M$ . Under hypothesis (1), $\dim_{k} M' = \dim_{k} M_{1}$ and $M_{1}$ occurs at least once, so the composition series of $M'$ consists of that single factor, whence $M' ≅ M_{1} = M$ .

Corollary(7.21)Splitting fields separate simple modules

Let $R$ be a $k$ -algebra with $\dim_{k} R < \infty$ which splits over $k$ . Then two simple left $R$ -modules are isomorphic if and only if they have the same character. This holds in every characteristic.

Indeed every simple module is absolutely irreducible by the definition of a splitting field, so $(7.20)$ applies with hypothesis (2).

CorollaryThe simple characters are a basis

Let $R$ split over $k$ , with simple left modules $M_{1}, \dots, M_{r}$ . Then $χ_{M_{1}}, \dots, χ_{M_{r}}$ are linearly independent in ${Hom}_{k} (R, k)$ , and they form a basis of the space of linear functionals vanishing on $T (R) = rad R + [R, R]$ .

Proof

Each $χ_{M_{i}}$ vanishes on $[R, R]$ because traces are symmetric, and on $rad R$ because the radical annihilates simple modules; so all $r$ functionals kill $T (R)$ .

For independence, fix $j$ and choose $a \in R$ projecting to a rank-one idempotent in $B_{j} ≅ M_{n_{j}} (k)$ and to $0$ in the other components. Then $χ_{M_{i}} (a) = δ_{i j}$ . Applying a hypothetical relation $\sum_{i} c_{i} χ_{M_{i}} = 0$ to this element gives $c_{j} = 0$ for each $j$ .

Finally, the space of functionals vanishing on $T (R)$ has dimension $\dim_{k} R / T (R)$ , which equals $r$ by $(7.17)$ since $R$ splits. A linearly independent family of the right size is a basis.

Proof Techniques and Method

How these proofs work, and which move to reuse.

Three moves, all variations on the same idea: manufacture an element of $R$ that acts in a prescribed way.

Move 1

Address one component

$R ↠ \prod_{i} B_{i}$ lets one choose an element supported in a single simple component. Evaluating a character there isolates exactly one multiplicity.

Move 2

Choose the trace, not just the support

Absolute irreducibility upgrades Move 1: because $R \to End (M_{1})_{k}$ is onto, one may prescribe the trace to be $1$ . This is what makes $(7.20)$ characteristic-free.

Move 3

Read an integer off a scalar

An equation in $k$ constrains an integer multiplicity only as far as the characteristic allows. In characteristic $0$ it determines it; in characteristic $p$ it determines it modulo $p$ , and $m_{1} \neq 0$ in $k$ still forces $m_{1} \geq 1$ .

Move 3 is the reason $(7.20)$ survives characteristic $p$ while $(7.19)$ does not: the conclusion " $m_{1} \geq 1$ " needs only that $m_{1}$ is nonzero in $k$ , whereas recovering all the $m_{i}$ needs division.

Worked Example

A split algebra in characteristic zero: $ℚ S_{3}$

$ℚ S_{3} ≅ ℚ \times ℚ \times M_{2} (ℚ)$ , so $ℚ$ is a splitting field and there are three simple modules: the trivial module, the sign module, and the two-dimensional standard module. Their characters, restricted to $G$ and evaluated on conjugacy class representatives, are:

Character values on the classes of $S_{3}$
Simple module	$\dim_{ℚ}$	$1$	transposition	$3$ -cycle
trivial	$1$	$1$	$1$	$1$
sign	$1$	$1$	$- 1$	$1$
standard	$2$	$2$	$0$	$- 1$

The three rows are linearly independent, as the corollary predicts. The commutator computation agrees: $[R, R] = 0 \times 0 \times 𝔰_{𝔩 2} (ℚ)$ has dimension $3$ and $rad R = 0$ , so $\dim_{ℚ} R / T (R) = 6 - 3 = 3$ — exactly the number of simple modules, and exactly the dimension of the space of trace functions the characters span.

Equal characters, non-isomorphic modules

Take $k = ℚ$ and $R = k [t] / (t^{2})$ . Consider

M : t \mapsto (\begin{matrix} 0 & 0 \\ 1 & 0 \end{matrix}), M' : t \mapsto (\begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix})

Both matrices have trace $0$ , so $χ_{M} (a + b t) = 2 a = χ_{M'} (a + b t)$ for all $a, b \in k$ . The modules have the same composition factors — two copies of the unique simple module $k$ — as $(7.19)$ guarantees. They are not isomorphic: $M$ is indecomposable, $M'$ is not. Characteristic $0$ does not rescue this; only semisimplicity would.

A characteristic $p$ collapse

Let $char k = p$ and $R = k C_{p}$ with $C_{p} = ⟨ g ⟩$ . Then $R ≅ k [x] / (x^{p})$ with $x = g - 1$ , so there is one simple module, the trivial module $k$ , and $rad R = (x)$ of dimension $p - 1$ .

The regular module $_{R} R$ has $χ (g^{j}) = 0$ for $j \neq 0$ , since $g^{j}$ permutes the basis ${1, g, \dots, g^{p - 1}}$ without fixed points, and $χ (1) = p \cdot 1_{k} = 0$ . So the character of the regular module is identically zero, and it does not even record $\dim_{k} R = p$ . Its composition factors are $p$ copies of the trivial module — invisible to the character.

Comparison and Classification

What the character determines
	$char k = 0$	$char k = p$	$k$ splits $R$
$\dim_{k} M$	yes	no	partial
Composition factors with multiplicity	yes	no	partial
Isomorphism class, $M$ semisimple	yes	no	partial
Isomorphism class, $M$ arbitrary	no	no	no
Isomorphism class, $M$ absolutely irreducible	yes	yes	yes
Linear independence of simple characters	yes	partial	yes

What the character determines

Hypotheses of the two theorems
	$(7.19)$	$(7.20)$
Characteristic	$0$ only	arbitrary
Modules	any finite-dimensional	$M$ absolutely irreducible
Extra condition	semisimplicity for isomorphism	equal dimensions, or $M'$ irreducible
Conclusion	composition factors with multiplicity	isomorphism
Key evaluation point	$a_{i}$ , identity on $M_{i}$	$a$ , trace $1$ on $M_{1}$

Relationship Map

Characters sit at the junction of the commutator subspace, the splitting theory and the classical theory of group representations.

χ_{M}

kills

[R, R]

trace functions on

R

basis when

R

splits

r = \dim_{k} R / T (R)

Character theory needs
- from this chapter
  - absolute irreducibility, to prescribe a trace (7.5)
  - the product decomposition of $R / rad R$ (7.1)
  - the commutator subspace, to count independent characters (7.16)
- and supplies
  - separation of simple modules over a splitting field (7.21)
  - the orthogonality relations in the semisimple case
  - the starting point for Brauer characters in characteristic $p$

Applications and Industry Use

Applications here means where this structure is used — inside mathematics and in the engineering and computing disciplines that consume it.

Finite group theory

Character tables

The entire classical theory — orthogonality, induction, Burnside's $p^{a} q^{b}$ theorem — rests on the fact that in characteristic $0$ a semisimple module is determined by its character.

Chemistry and physics

Symmetry analysis

Decomposing a vibrational or orbital representation into irreducibles is done by evaluating a character on conjugacy classes and solving a small linear system. That method is exactly $(7.19)$ .

Modular representation theory

Brauer characters

In characteristic $p$ ordinary characters lose multiplicities. Brauer characters restore the theory by evaluating on $p$ -regular elements and lifting eigenvalues to characteristic $0$ .

Computational algebra

Cheap module fingerprints

Comparing characters is far cheaper than testing module isomorphism. Systems use character comparison as a fast necessary condition, then invoke an isomorphism test only when characters agree.

The recurring engineering value is compression: a module of dimension $n$ is summarised by $\dim_{k} R$ scalars, and over a splitting field in characteristic $0$ nothing that matters is lost.

Standards and Notation

Standards here covers notation, symbol and markup standards, and reference implementations, rather than material or design codes.

Notation

χ_{M}

for the character of

M

;

χ_{ρ}

when a representation

ρ

is named

Group case

χ (g)

for

g \in G

; class functions are tabulated on conjugacy classes

Trace

tr

is the ordinary trace of a

k

-linear map, not a reduced trace

Modular caseBrauer characters are functions on

p

-regular classes with values in a characteristic zero field

GAPIrr, BrauerCharacterValue, CharacterTable

MagmaCharacterTable, BrauerCharacter

MarkupPresentation MathML per ISO/IEC 40314; symbol conventions per ISO 80000-2

Failure Modes and Common Mistakes

Do not assume $χ_{M} (1) = \dim_{k} M$ as an integer; it is $\dim_{k} M \cdot 1_{k}$ , which vanishes when the characteristic divides the dimension.
Do not use characters to detect extensions; they are constant on Grothendieck classes by construction.
Do not apply $(7.21)$ without checking that $k$ splits $R$ — over $ℝ$ two non-isomorphic simple modules can be forced apart only after extending the field.

Quick Reference

Definition

χ_{M} (a) = tr (a ∣ M)

, a

k

-linear functional

Symmetry

χ_{M} (a b) = χ_{M} (b a)

, so

χ_{M}

kills

[R, R]

Value at $1$

(\dim_{k} M) \cdot 1_{k}

Additivity

χ_{M} = χ_{M'} + χ_{M''}

on short exact sequences

(7.19)

char k = 0

: composition factors and multiplicities recovered

(7.19), semisimpleequal characters imply isomorphic

(7.20)

M

absolutely irreducible plus equal dimensions or

M'

simple

(7.21)

R

split: simple modules separated by characters

Basisthe

r

simple characters span the functionals killing

T (R)

Statement finder
Result	Content	Reference
Definition and additivity	trace functional, additive on exact sequences	before (7.19)
Characteristic zero	composition factors determined	(7.19)
Non-semisimple failure	$k [t] / (t^{2})$ , two modules of dimension $2$	after (7.19)
Characteristic $p$ failure	$p$ copies of a simple module	after (7.19)
Separation	absolutely irreducible modules determined by characters	(7.20)
Split algebras	simple modules separated in any characteristic	(7.21)

Frequently Asked Questions

Why does the character only see composition factors?

Because trace is additive along block-triangular decompositions. Choosing a basis adapted to a submodule makes every action block upper triangular, and the trace is the sum of the diagonal blocks' traces. Extension data lives in the off-diagonal block, which the trace ignores.

Is a character determined by its values on a basis of

R

Yes, since it is $k$ -linear. For a group algebra it is determined by its values on group elements, and in fact only by the values on conjugacy class representatives, since it vanishes on all commutators.

Why is (7.20) true in characteristic

p

when (7.19) is not?

Because it needs only that $m_{1}$ is nonzero in $k$ , which forces $m_{1} \geq 1$ as an integer. (7.19) needs to divide by $\dim_{k} M_{i}$ , which may be zero in $k$ . The absolute irreducibility of $M$ is what lets one prescribe the trace to be $1$ rather than $\dim_{k} M_{1}$ .

What replaces (7.19) in characteristic

p

Brauer character theory. One evaluates on $p$ -regular elements, lifts the eigenvalues of the action to roots of unity in characteristic $0$ , and sums them there. The resulting functions do determine composition factors of modules in characteristic $p$ .

Do characters of non-isomorphic simple modules have to differ?

Over a splitting field yes, by (7.21), in any characteristic. Over a general field they can coincide only if the modules fail to be absolutely irreducible, so the first place to look for pathologies is a field over which the algebra does not split.

How many independent characters can an algebra have?

At most $\dim_{k} R / [R, R]$ , since every character is a trace function. When $R$ splits the simple characters already achieve $\dim_{k} R / T (R) = r$ , and they form a basis of the functionals annihilating $T (R)$ .

References

T. Y. Lam, A First Course in Noncommutative Rings, Graduate Texts in Mathematics 131, Springer-Verlag, 1991, §7 (pp. 121–122).
C. W. Curtis and I. Reiner, Representation Theory of Finite Groups and Associative Algebras, Wiley-Interscience, 1962, §§30 and 82.
J.-P. Serre, Linear Representations of Finite Groups, Graduate Texts in Mathematics 42, Springer-Verlag, 1977, Part I and Part III.
W. Feit, The Representation Theory of Finite Groups, North-Holland, 1982, Chapter I.
I. M. Isaacs, Character Theory of Finite Groups, Academic Press, 1976, Chapters 2–3.

AI Suggested Questions

Define Brauer characters precisely and show they determine composition factors in characteristic $p$ .
Prove the first orthogonality relation from the results on this page in the semisimple split case.
Give two non-isomorphic modules over $𝔽_{3} S_{3}$ with the same character and different dimensions.
How is the character of an induced module computed from the character of the original module?
Show that the character of a projective module over $k G$ in characteristic $p$ vanishes on $p$ -singular elements.
What is the relation between the space of trace functions and the zeroth Hochschild homology of $R$ ?
How do computer algebra systems test module isomorphism when characters agree?

Characters of Algebras

Executive Summary

Overview

Learning Objectives

Definitions

Core Concepts

Additivity, and what it costs

Isolating a multiplicity

Key Results

Proof Techniques and Method

Address one component

Choose the trace, not just the support

Read an integer off a scalar

Worked Example

A split algebra in characteristic zero: $ℚ S_{3}$

Equal characters, non-isomorphic modules

A characteristic $p$ collapse

Comparison and Classification

Relationship Map

Applications and Industry Use

Character tables

Symmetry analysis

Brauer characters

Cheap module fingerprints

Standards and Notation

Failure Modes and Common Mistakes

Quick Reference

Frequently Asked Questions

References

AI Suggested Questions

Continue learning

01Executive Summary

02Overview

03Learning Objectives

04Definitions

05Core Concepts

Additivity, and what it costs

Isolating a multiplicity

06Key Results

07Proof Techniques and Method

Address one component

Choose the trace, not just the support

Read an integer off a scalar

08Worked Example

A split algebra in characteristic zero: ℚS3

Equal characters, non-isomorphic modules

A characteristic p collapse

09Comparison and Classification

10Relationship Map

11Applications and Industry Use

Character tables

Symmetry analysis

Brauer characters

Cheap module fingerprints

12Standards and Notation

13Failure Modes and Common Mistakes

14Quick Reference

15Frequently Asked Questions

16Related KEVOS Topics

17References

18AI Suggested Questions

Continue learning

Executive Summary

Overview

Learning Objectives

Definitions

Core Concepts

Key Results

Proof Techniques and Method

Worked Example

A split algebra in characteristic zero: $ℚ S_{3}$

A characteristic $p$ collapse

Comparison and Classification

Relationship Map

Applications and Industry Use

Standards and Notation

Failure Modes and Common Mistakes

Quick Reference

Frequently Asked Questions

Related KEVOS Topics

References

AI Suggested Questions