Proposition (D& F, Thm 10.5.33)

Let Q be an R-module. The following are equivalent.

(1) If 0 ãÑ AϕÝÑ B

ψÝÑ C Ñ 0 is exact, then so is

0 ãÑ HomRpC,Qqϕ1ÝÑ HomRpB,Qq

ψ1ÝÑ HomRpA,Qq Ñ 0.

(2) For any injective ϕ : AÑ B and f P HomRpA,Qq, there is alift F of f in HomRpB,Qq:

Q

A Bϕ

fF

(3) If Q is isomorphic to a submodule of B, then it is a direct

summand of B, i.e. every SES 0 ãÑ QϕÝÑ B

ψÝÑ C Ñ 0 splits.

(Compare to [D& F, Thm 10.5.29].)An R-module is called injective if it satisfies any of 1–3.Essentially, property 3 is our favorite again; it says that any moduleB that Q injects into has Q as a direct summand.

2. For any injective ϕ : AÑ B and f P HomRpA,Qq, there is alift F of f in HomRpB,Qq:

Q

A Bϕ

fF

Proposition (Baer’s Criterion)

An R-module Q is projective if and only if for every left ideal I ofR, every R-module homomorphism I Ñ Q lifts to an R-modulehomomorphism RÑ Q.

(See [D&F, Prop. 10.5.36(1)] for proof.)

Defn. An R-module M is divisible if for all r P R, we haverM “M .

Ex. Z is not divisible as a Z-module since nZ ‰ Z for n ‰ ˘1.

Ex. F3 is divisible as a F3-module.

Ex. Every field F is a divisible module over itself:Fˆ “ F´ 0 is a multiplicative group and therefore acts transitivelyon itself by left multiplication.

Proposition (D&F, Prop. 10.5.36(2))

If R is a PID, an R-module is injective if and only if it is divisible.

Ex. Z is not divisible over itself, so it is not injective.(But again,

0 ãÑ Z ˚nÝÑ Z π

ÝÑ Z{nZÑ 0

does not split.)Ex. Q is divisible over Z, so Q is an injective Z-module.

Defn. An R-module M is divisible if for all r P R, we haverM “M .

Ex. Z is not divisible as a Z-module since nZ ‰ Z for n ‰ ˘1.

Ex. F3 is divisible as a F3-module.

Ex. Every field F is a divisible module over itself:Fˆ “ F´ 0 is a multiplicative group and therefore acts transitivelyon itself by left multiplication.

Proposition (D&F, Prop. 10.5.36(2))

If R is a PID, an R-module is injective if and only if it is divisible.

Ex. Z is not divisible over itself, so it is not injective.(But again,

0 ãÑ Z ˚nÝÑ Z π

ÝÑ Z{nZÑ 0

does not split.)Ex. Q is divisible over Z, so Q is an injective Z-module.

Defn. An R-module M is divisible if for all r P R, we haverM “M .

Ex. Z is not divisible as a Z-module since nZ ‰ Z for n ‰ ˘1.

Ex. F3 is divisible as a F3-module.

Ex. Every field F is a divisible module over itself:Fˆ “ F´ 0 is a multiplicative group and therefore acts transitivelyon itself by left multiplication.

Proposition (D&F, Prop. 10.5.36(2))

If R is a PID, an R-module is injective if and only if it is divisible.

Ex. Z is not divisible over itself, so it is not injective.(But again,

0 ãÑ Z ˚nÝÑ Z π

ÝÑ Z{nZÑ 0

does not split.)Ex. Q is divisible over Z, so Q is an injective Z-module.

Defn. An R-module M is divisible if for all r P R, we haverM “M .

Ex. Z is not divisible as a Z-module since nZ ‰ Z for n ‰ ˘1.

Ex. F3 is divisible as a F3-module.

Ex. Every field F is a divisible module over itself:Fˆ “ F´ 0 is a multiplicative group and therefore acts transitivelyon itself by left multiplication.

Proposition (D&F, Prop. 10.5.36(2))

If R is a PID, an R-module is injective if and only if it is divisible.

Ex. Z is not divisible over itself, so it is not injective.(But again,

0 ãÑ Z ˚nÝÑ Z π

ÝÑ Z{nZÑ 0

does not split.)Ex. Q is divisible over Z, so Q is an injective Z-module.

Defn. An R-module M is divisible if for all r P R, we haverM “M .

Ex. Z is not divisible as a Z-module since nZ ‰ Z for n ‰ ˘1.

Ex. F3 is divisible as a F3-module.

Ex. Every field F is a divisible module over itself:Fˆ “ F´ 0 is a multiplicative group and therefore acts transitivelyon itself by left multiplication.

Proposition (D&F, Prop. 10.5.36(2))

If R is a PID, an R-module is injective if and only if it is divisible.

Ex. Z is not divisible over itself, so it is not injective.

(But again,

0 ãÑ Z ˚nÝÑ Z π

ÝÑ Z{nZÑ 0

does not split.)Ex. Q is divisible over Z, so Q is an injective Z-module.

Defn. An R-module M is divisible if for all r P R, we haverM “M .

Ex. Z is not divisible as a Z-module since nZ ‰ Z for n ‰ ˘1.

Ex. F3 is divisible as a F3-module.

Ex. Every field F is a divisible module over itself:Fˆ “ F´ 0 is a multiplicative group and therefore acts transitivelyon itself by left multiplication.

Proposition (D&F, Prop. 10.5.36(2))

If R is a PID, an R-module is injective if and only if it is divisible.

Ex. Z is not divisible over itself, so it is not injective.(But again,

0 ãÑ Z ˚nÝÑ Z π

ÝÑ Z{nZÑ 0

does not split.)

Ex. Q is divisible over Z, so Q is an injective Z-module.

Defn. An R-module M is divisible if for all r P R, we haverM “M .

Ex. Z is not divisible as a Z-module since nZ ‰ Z for n ‰ ˘1.

Ex. F3 is divisible as a F3-module.

Ex. Every field F is a divisible module over itself:Fˆ “ F´ 0 is a multiplicative group and therefore acts transitivelyon itself by left multiplication.

Proposition (D&F, Prop. 10.5.36(2))

If R is a PID, an R-module is injective if and only if it is divisible.

Ex. Z is not divisible over itself, so it is not injective.(But again,

0 ãÑ Z ˚nÝÑ Z π

ÝÑ Z{nZÑ 0

does not split.)Ex. Q is divisible over Z, so Q is an injective Z-module.

Theorem (D&F, Thm. 10.5.38)

Every R-module M is contained in an injective R-module.Moreover, the minimal such Q is well-defined.

Minimality: Fix M . There exists an injective Q such that

for all injective Q1 ĚM , we have Q – Q for some M Ď Q Ď Q1

i.e. up to isomorphism M Ď Q Ď Q1.

We call Q the injective hull or injective envelope of M .

Semisimple rings: decompositions galore

A ring R is semisimple if every R module is isomorphic to thedirect sum of simple modules. Equivalently, R is semisimple if R isa direct sum of simple rings.

Non-ex: R “ Z. The let regular module M “ Z is not simple (ithas submodules nZ for all n), but does not split. So Z is notsemisimple.

Example: We will show FG is semisimple for all finite groups Gand fields F whose characteristic of F does not divide |G|. Forexample, CSn is semisimple because C has characteristic 0.

Non-ex: F3 has characteristic 3, which divides |S3| “ 6; and F3S3is not semisimple. For example, the permutation module is notsimple but does not split.

Semisimple rings: decompositions galore

A ring R is semisimple if every R module is isomorphic to thedirect sum of simple modules. Equivalently, R is semisimple if R isa direct sum of simple rings.

Non-ex: R “ Z. The let regular module M “ Z is not simple (ithas submodules nZ for all n), but does not split. So Z is notsemisimple.

Example: We will show FG is semisimple for all finite groups Gand fields F whose characteristic of F does not divide |G|. Forexample, CSn is semisimple because C has characteristic 0.

Non-ex: F3 has characteristic 3, which divides |S3| “ 6; and F3S3is not semisimple. For example, the permutation module is notsimple but does not split.

Semisimple rings: decompositions galore

A ring R is semisimple if every R module is isomorphic to thedirect sum of simple modules. Equivalently, R is semisimple if R isa direct sum of simple rings.

Non-ex: R “ Z. The let regular module M “ Z is not simple (ithas submodules nZ for all n), but does not split. So Z is notsemisimple.

Example: We will show FG is semisimple for all finite groups Gand fields F whose characteristic of F does not divide |G|. Forexample, CSn is semisimple because C has characteristic 0.

Non-ex: F3 has characteristic 3, which divides |S3| “ 6; and F3S3is not semisimple. For example, the permutation module is notsimple but does not split.

Semisimple rings: decompositions galore

A ring R is semisimple if every R module is isomorphic to thedirect sum of simple modules. Equivalently, R is semisimple if R isa direct sum of simple rings.

Non-ex: R “ Z. The let regular module M “ Z is not simple (ithas submodules nZ for all n), but does not split. So Z is notsemisimple.

Example: We will show FG is semisimple for all finite groups Gand fields F whose characteristic of F does not divide |G|. Forexample, CSn is semisimple because C has characteristic 0.

Non-ex: F3 has characteristic 3, which divides |S3| “ 6; and F3S3is not semisimple. For example, the permutation module is notsimple but does not split.

Theorem (Maschke’s theorem [D&F, Thm. 18.1.1])

Let G be a finite group and let F be a field whose characteristicdoes not divide |G|. Let V be an FG-module, and U be asubmodule.

Then there is also a submodule W satisfying V – U ‘W .(There’s a submodule W in V satisfying U XW “ 0 and U `W “ V .)

Before we get to the proof. . .

Big example:Let

G be a group of order n,F be a field of char not dividing n,and consider the module V “ FG over R “ FG.

Given the linear algebra that we know, we can decompose FGusing inner products and projections as follows. . .

Theorem (Maschke’s theorem [D&F, Thm. 18.1.1])

Let G be a finite group and let F be a field whose characteristicdoes not divide |G|. Let V be an FG-module, and U be asubmodule.

Then there is also a submodule W satisfying V – U ‘W .(There’s a submodule W in V satisfying U XW “ 0 and U `W “ V .)

Before we get to the proof. . .

Big example:Let

G be a group of order n,F be a field of char not dividing n,and consider the module V “ FG over R “ FG.

Given the linear algebra that we know, we can decompose FGusing inner products and projections as follows. . .

Theorem (Maschke’s theorem [D&F, Thm. 18.1.1])

Let G be a finite group and let F be a field whose characteristicdoes not divide |G|. Let V be an FG-module, and U be asubmodule.

Then there is also a submodule W satisfying V – U ‘W .(There’s a submodule W in V satisfying U XW “ 0 and U `W “ V .)

Before we get to the proof. . .

Big example:Let

G be a group of order n,F be a field of char not dividing n,and consider the module V “ FG over R “ FG.

Given the linear algebra that we know, we can decompose FGusing inner products and projections as follows. . .

Big example: every submodule of FG is a direct summand

LetG be a group of order n,F be a field of char not dividing n,and consider the module V “ FG over R “ FG.

Define the sym. bilinear form x, y on FG by

xg, hy “ δg,h for all g, h P G,

and extend linearly.

Claim 1: x, y is non-degenerate.If a “

ř

gPG αgg satisfies

xa, FGy “ 0, then xa, hy “ 0 for all h P G.

So

0 “

C

ÿ

gPG

αgg, h

G

“ÿ

gPG

αgxg, hy “ αh

for all h P G. So a “ 0. X

Big example: every submodule of FG is a direct summand

LetG be a group of order n,F be a field of char not dividing n,and consider the module V “ FG over R “ FG.

Define the sym. bilinear form x, y on FG by

xg, hy “ δg,h for all g, h P G,

and extend linearly.

Claim 1: x, y is non-degenerate.

If a “ř

gPG αgg satisfies

xa, FGy “ 0, then xa, hy “ 0 for all h P G.

So

0 “

C

ÿ

gPG

αgg, h

G

“ÿ

gPG

αgxg, hy “ αh

for all h P G. So a “ 0. X

Big example: every submodule of FG is a direct summand

LetG be a group of order n,F be a field of char not dividing n,and consider the module V “ FG over R “ FG.

Define the sym. bilinear form x, y on FG by

xg, hy “ δg,h for all g, h P G,

and extend linearly.

Claim 1: x, y is non-degenerate.If a “

ř

gPG αgg satisfies

xa, FGy “ 0, then xa, hy “ 0 for all h P G.

So

0 “

C

ÿ

gPG

αgg, h

G

“ÿ

gPG

αgxg, hy “ αh

for all h P G. So a “ 0. X

Big example: every submodule of FG is a direct summand

LetG be a group of order n,F be a field of char not dividing n,and consider the module V “ FG over R “ FG.

Define the sym. bilinear form x, y on FG by

xg, hy “ δg,h for all g, h P G,

and extend linearly.

Claim 1: x, y is non-degenerate.If a “

ř

gPG αgg satisfies

xa, FGy “ 0, then xa, hy “ 0 for all h P G.

So

0 “

C

ÿ

gPG

αgg, h

G

“ÿ

gPG

αgxg, hy “ αh

for all h P G. So a “ 0. X

Big example: every submodule of FG is a direct summand

LetG be a group of order n,F be a field of char not dividing n,and consider the module V “ FG over R “ FG.

Define the sym. bilinear form x, y on FG by

xg, hy “ δg,h for all g, h P G,

and extend linearly.

Claim 1: x, y is non-degenerate.If a “

ř

gPG αgg satisfies

xa, FGy “ 0, then xa, hy “ 0 for all h P G.

So

0 “

C

ÿ

gPG

αgg, h

G

“ÿ

gPG

αgxg, hy

“ αh

for all h P G. So a “ 0. X

Big example: every submodule of FG is a direct summand

LetG be a group of order n,F be a field of char not dividing n,and consider the module V “ FG over R “ FG.

Define the sym. bilinear form x, y on FG by

xg, hy “ δg,h for all g, h P G,

and extend linearly.

Claim 1: x, y is non-degenerate.If a “

ř

gPG αgg satisfies

xa, FGy “ 0, then xa, hy “ 0 for all h P G.

So

0 “

C

ÿ

gPG

αgg, h

G

“ÿ

gPG

αgxg, hy “ αh

for all h P G. So a “ 0. X

Big example: every submodule of FG is a direct summandLet

G be a group of order n,F be a field of char not dividing n,and consider the module V “ FG over R “ FG.

Define the sym. bilinear form x, y on FG by

xg, hy “ δg,h for all g, h P G,

and extend linearly.

Claim 2: x, y is invariant. (meaning xgu, vy “ xu, g´1vy @g, u, v)

Fix an order on G “ tg1, . . . , gnu, yielding an ordered basis for FGas a vector space.

Since every g P G Ď R acts on G Ď V by apermutation, the associated matrix M “MG

G pρpgqq for the actionof g on FG is a permutation matrix.(Ch4: G acts on itself by left mult., which induces a hom into SG.)

In particular, M´1 “M t. So, identifying a P FG with M1Bpaq, we

have

xga, by “ pMaqtb “ atM tb “ atM´1b “ xa, g´1by

for all a, b P FG. X

Big example: every submodule of FG is a direct summandLet

G be a group of order n,F be a field of char not dividing n,and consider the module V “ FG over R “ FG.

Define the sym. bilinear form x, y on FG by

xg, hy “ δg,h for all g, h P G,

and extend linearly.

Claim 2: x, y is invariant. (meaning xgu, vy “ xu, g´1vy @g, u, v)

Fix an order on G “ tg1, . . . , gnu, yielding an ordered basis for FGas a vector space. Since every g P G Ď R acts on G Ď V by apermutation, the associated matrix M “MG

G pρpgqq for the actionof g on FG is a permutation matrix.(Ch4: G acts on itself by left mult., which induces a hom into SG.)

In particular, M´1 “M t. So, identifying a P FG with M1Bpaq, we

have

xga, by “ pMaqtb “ atM tb “ atM´1b “ xa, g´1by

for all a, b P FG. X

Big example: every submodule of FG is a direct summandLet

G be a group of order n,F be a field of char not dividing n,and consider the module V “ FG over R “ FG.

Define the sym. bilinear form x, y on FG by

xg, hy “ δg,h for all g, h P G,

and extend linearly.

Claim 2: x, y is invariant. (meaning xgu, vy “ xu, g´1vy @g, u, v)

Fix an order on G “ tg1, . . . , gnu, yielding an ordered basis for FGas a vector space. Since every g P G Ď R acts on G Ď V by apermutation, the associated matrix M “MG

G pρpgqq for the actionof g on FG is a permutation matrix.(Ch4: G acts on itself by left mult., which induces a hom into SG.)

In particular, M´1 “M t.

So, identifying a P FG with M1Bpaq, we

have

xga, by “ pMaqtb “ atM tb “ atM´1b “ xa, g´1by

for all a, b P FG. X

Big example: every submodule of FG is a direct summandLet

G be a group of order n,F be a field of char not dividing n,and consider the module V “ FG over R “ FG.

Define the sym. bilinear form x, y on FG by

xg, hy “ δg,h for all g, h P G,

and extend linearly.

Claim 2: x, y is invariant. (meaning xgu, vy “ xu, g´1vy @g, u, v)

Fix an order on G “ tg1, . . . , gnu, yielding an ordered basis for FGas a vector space. Since every g P G Ď R acts on G Ď V by apermutation, the associated matrix M “MG

G pρpgqq for the actionof g on FG is a permutation matrix.(Ch4: G acts on itself by left mult., which induces a hom into SG.)

In particular, M´1 “M t. So, identifying a P FG with M1Bpaq, we

have

xga, by “ pMaqtb “ atM tb

“ atM´1b “ xa, g´1by

for all a, b P FG. X

Big example: every submodule of FG is a direct summandLet

G be a group of order n,F be a field of char not dividing n,and consider the module V “ FG over R “ FG.

Define the sym. bilinear form x, y on FG by

xg, hy “ δg,h for all g, h P G,

and extend linearly.

Claim 2: x, y is invariant. (meaning xgu, vy “ xu, g´1vy @g, u, v)

Fix an order on G “ tg1, . . . , gnu, yielding an ordered basis for FGas a vector space. Since every g P G Ď R acts on G Ď V by apermutation, the associated matrix M “MG

G pρpgqq for the actionof g on FG is a permutation matrix.(Ch4: G acts on itself by left mult., which induces a hom into SG.)

In particular, M´1 “M t. So, identifying a P FG with M1Bpaq, we

have

xga, by “ pMaqtb “ atM tb “ atM´1b “ xa, g´1by

for all a, b P FG. X

Big example: every submodule of FG is a direct summand

Let G grp of order n, F field of char - n, and consider V “ FGover R “ FG. Fix order G “ tg1, . . . , gnu, and define x, y on FGby xg, hy “ δg,h and extending linearly. Then x, y is symmetric,bilinear, invariant, and nondegenerate.

Let U be a submodule of FG, and define

UK “ tv P FG | xu, vy “ 0u.

Nondegenerate ñ dimpUKq “ dimpFGq ´ dimpUq.Bilinear ñ UK is a subspace.Invariant ñ UK is closed under the FG-action:If xu, vy “ 0 for all u P U , then for all g P G,

xu, gvy “ xg´1u, vy “ 0, since U is closed under the G-action.

So gv P UK for all g P G, and thus´

ř

gPG αg¯

v P UK. X

Remains to show: U ` UK “ FG and U X UK “ 0.Note: The first implies second by a dimension count.

Big example: every submodule of FG is a direct summand

Let G grp of order n, F field of char - n, and consider V “ FGover R “ FG. Fix order G “ tg1, . . . , gnu, and define x, y on FGby xg, hy “ δg,h and extending linearly. Then x, y is symmetric,bilinear, invariant, and nondegenerate.

Let U be a submodule of FG, and define

UK “ tv P FG | xu, vy “ 0u.

Nondegenerate ñ dimpUKq “ dimpFGq ´ dimpUq.Bilinear ñ UK is a subspace.Invariant ñ UK is closed under the FG-action:If xu, vy “ 0 for all u P U , then for all g P G,

xu, gvy “ xg´1u, vy “ 0, since U is closed under the G-action.

So gv P UK for all g P G, and thus´

ř

gPG αg¯

v P UK. X

Remains to show: U ` UK “ FG and U X UK “ 0.Note: The first implies second by a dimension count.

Big example: every submodule of FG is a direct summand

Let G grp of order n, F field of char - n, and consider V “ FGover R “ FG. Fix order G “ tg1, . . . , gnu, and define x, y on FGby xg, hy “ δg,h and extending linearly. Then x, y is symmetric,bilinear, invariant, and nondegenerate.

Let U be a submodule of FG, and define

UK “ tv P FG | xu, vy “ 0u.

Nondegenerate ñ dimpUKq “ dimpFGq ´ dimpUq.

Bilinear ñ UK is a subspace.Invariant ñ UK is closed under the FG-action:If xu, vy “ 0 for all u P U , then for all g P G,

xu, gvy “ xg´1u, vy “ 0, since U is closed under the G-action.

So gv P UK for all g P G, and thus´

ř

gPG αg¯

v P UK. X

Remains to show: U ` UK “ FG and U X UK “ 0.Note: The first implies second by a dimension count.

Big example: every submodule of FG is a direct summand

Let G grp of order n, F field of char - n, and consider V “ FGover R “ FG. Fix order G “ tg1, . . . , gnu, and define x, y on FGby xg, hy “ δg,h and extending linearly. Then x, y is symmetric,bilinear, invariant, and nondegenerate.

Let U be a submodule of FG, and define

UK “ tv P FG | xu, vy “ 0u.

Nondegenerate ñ dimpUKq “ dimpFGq ´ dimpUq.Bilinear ñ UK is a subspace.

Invariant ñ UK is closed under the FG-action:If xu, vy “ 0 for all u P U , then for all g P G,

xu, gvy “ xg´1u, vy “ 0, since U is closed under the G-action.

So gv P UK for all g P G, and thus´

ř

gPG αg¯

v P UK. X

Remains to show: U ` UK “ FG and U X UK “ 0.Note: The first implies second by a dimension count.

Big example: every submodule of FG is a direct summand

Let G grp of order n, F field of char - n, and consider V “ FGover R “ FG. Fix order G “ tg1, . . . , gnu, and define x, y on FGby xg, hy “ δg,h and extending linearly. Then x, y is symmetric,bilinear, invariant, and nondegenerate.

Let U be a submodule of FG, and define

UK “ tv P FG | xu, vy “ 0u.

Nondegenerate ñ dimpUKq “ dimpFGq ´ dimpUq.Bilinear ñ UK is a subspace.Invariant ñ UK is closed under the FG-action:

If xu, vy “ 0 for all u P U , then for all g P G,

xu, gvy “ xg´1u, vy “ 0, since U is closed under the G-action.

So gv P UK for all g P G, and thus´

ř

gPG αg¯

v P UK. X

Remains to show: U ` UK “ FG and U X UK “ 0.Note: The first implies second by a dimension count.

Big example: every submodule of FG is a direct summand

Let G grp of order n, F field of char - n, and consider V “ FGover R “ FG. Fix order G “ tg1, . . . , gnu, and define x, y on FGby xg, hy “ δg,h and extending linearly. Then x, y is symmetric,bilinear, invariant, and nondegenerate.

Let U be a submodule of FG, and define

UK “ tv P FG | xu, vy “ 0u.

Nondegenerate ñ dimpUKq “ dimpFGq ´ dimpUq.Bilinear ñ UK is a subspace.Invariant ñ UK is closed under the FG-action:If xu, vy “ 0 for all u P U , then for all g P G,

xu, gvy “ xg´1u, vy “ 0, since U is closed under the G-action.

So gv P UK for all g P G, and thus´

ř

gPG αg¯

v P UK. X

Remains to show: U ` UK “ FG and U X UK “ 0.Note: The first implies second by a dimension count.

Big example: every submodule of FG is a direct summand

Let G grp of order n, F field of char - n, and consider V “ FGover R “ FG. Fix order G “ tg1, . . . , gnu, and define x, y on FGby xg, hy “ δg,h and extending linearly. Then x, y is symmetric,bilinear, invariant, and nondegenerate.

Let U be a submodule of FG, and define

UK “ tv P FG | xu, vy “ 0u.

Nondegenerate ñ dimpUKq “ dimpFGq ´ dimpUq.Bilinear ñ UK is a subspace.Invariant ñ UK is closed under the FG-action:If xu, vy “ 0 for all u P U , then for all g P G,

xu, gvy “ xg´1u, vy “ 0, since U is closed under the G-action.

So gv P UK for all g P G, and thus´

ř

gPG αg¯

v P UK. X

Remains to show: U ` UK “ FG and U X UK “ 0.Note: The first implies second by a dimension count.

Big example: every submodule of FG is a direct summand

Let G grp of order n, F field of char - n, and consider V “ FGover R “ FG. Fix order G “ tg1, . . . , gnu, and define x, y on FGby xg, hy “ δg,h and extending linearly. Then x, y is symmetric,bilinear, invariant, and nondegenerate.

Let U be a submodule of FG, and define

UK “ tv P FG | xu, vy “ 0u.

Nondegenerate ñ dimpUKq “ dimpFGq ´ dimpUq.Bilinear ñ UK is a subspace.Invariant ñ UK is closed under the FG-action:If xu, vy “ 0 for all u P U , then for all g P G,

xu, gvy “ xg´1u, vy “ 0, since U is closed under the G-action.

So gv P UK for all g P G, and thus´

ř

gPG αg¯

v P UK. X

Remains to show: U ` UK “ FG and U X UK “ 0.Note: The first implies second by a dimension count.

Let G grp of order n, F field of char - n, and consider V “ FGover R “ FG. Define x, y on FG by xg, hy “ δg,h and extendinglinearly.

Showing U ` UK “ FG:Recall, for any a, b P V , we can write a “ b{{ ` bK where

b{{ “xa, by

xb, byb P Fb and bK “ a´ b{{ P pFbq

K,

since

xbK, by “ xa, by ´xa, by

xb, byxb, by “ 0.

Similarly, in any vector space V “ Fn with subspace U , any vectorv can be decomposed as v “ u` uK where u P U and uK P U

K asfollows. . .

Let G grp of order n, F field of char - n, and consider V “ FGover R “ FG. Define x, y on FG by xg, hy “ δg,h and extendinglinearly.

Showing U ` UK “ FG:Recall, for any a, b P V , we can write a “ b{{ ` bK where

b{{ “xa, by

xb, byb P Fb and bK “ a´ b{{ P pFbq

K,

since

xbK, by “ xa, by ´xa, by

xb, byxb, by “ 0.

Similarly, in any vector space V “ Fn with subspace U , any vectorv can be decomposed as v “ u` uK where u P U and uK P U

K asfollows. . .

Let G grp of order n, F field of char - n, and consider V “ FGover R “ FG. Define x, y on FG by xg, hy “ δg,h and extendinglinearly.

Showing U ` UK “ V :Goal: for each v P V , find u P U and uK P U

K such thatv “ u` uK.

Let B “ tu1, . . . , umu be a basis of U , and write

u “ α1u1 ` ¨ ¨ ¨ ` αmum for some αi P F.

Solving v ´ u P UK is the same as solving

0 “ xui, v ´ uy “ xui, vy ` α1xui, u1y ` ¨ ¨ ¨ ` αmxui, umy

for all i “ 1, . . . ,m. But since x, y is the dot product on V , this isthe same as finding α “ pα1, . . . , αmq

t such that

0 “M tpv ´Mαq, where M “

¨

˝

| |

u1 ¨ ¨ ¨ um| |

˛

‚PMn,mpF q.

Let G grp of order n, F field of char - n, and consider V “ FGover R “ FG. Define x, y on FG by xg, hy “ δg,h and extendinglinearly.

Showing U ` UK “ V :Goal: for each v P V , find u P U and uK P U

K such thatv “ u` uK. Let B “ tu1, . . . , umu be a basis of U , and write

u “ α1u1 ` ¨ ¨ ¨ ` αmum for some αi P F.

Solving v ´ u P UK is the same as solving

0 “ xui, v ´ uy “ xui, vy ` α1xui, u1y ` ¨ ¨ ¨ ` αmxui, umy

for all i “ 1, . . . ,m. But since x, y is the dot product on V , this isthe same as finding α “ pα1, . . . , αmq

t such that

0 “M tpv ´Mαq, where M “

¨

˝

| |

u1 ¨ ¨ ¨ um| |

˛

‚PMn,mpF q.

Let G grp of order n, F field of char - n, and consider V “ FGover R “ FG. Define x, y on FG by xg, hy “ δg,h and extendinglinearly.

Showing U ` UK “ V :Goal: for each v P V , find u P U and uK P U

K such thatv “ u` uK. Let B “ tu1, . . . , umu be a basis of U , and write

u “ α1u1 ` ¨ ¨ ¨ ` αmum for some αi P F.

Solving v ´ u P UK is the same as solving

0 “ xui, v ´ uy “ xui, vy ` α1xui, u1y ` ¨ ¨ ¨ ` αmxui, umy

for all i “ 1, . . . ,m.

But since x, y is the dot product on V , this isthe same as finding α “ pα1, . . . , αmq

t such that

0 “M tpv ´Mαq, where M “

¨

˝

| |

u1 ¨ ¨ ¨ um| |

˛

‚PMn,mpF q.

Let G grp of order n, F field of char - n, and consider V “ FGover R “ FG. Define x, y on FG by xg, hy “ δg,h and extendinglinearly.

Showing U ` UK “ V :Goal: for each v P V , find u P U and uK P U

K such thatv “ u` uK. Let B “ tu1, . . . , umu be a basis of U , and write

u “ α1u1 ` ¨ ¨ ¨ ` αmum for some αi P F.

Solving v ´ u P UK is the same as solving

0 “ xui, v ´ uy “ xui, vy ` α1xui, u1y ` ¨ ¨ ¨ ` αmxui, umy

for all i “ 1, . . . ,m. But since x, y is the dot product on V , this isthe same as finding α “ pα1, . . . , αmq

t such that

0 “M tpv ´Mαq, where M “

¨

˝

| |

u1 ¨ ¨ ¨ um| |

˛

‚PMn,mpF q.

Let G grp of order n, F field of char - n, and consider V “ FGover R “ FG. Define x, y on FG by xg, hy “ δg,h and extendinglinearly.

Showing U ` UK “ V :Goal: for each v P V , find u P U and uK P U

K such thatv “ u` uK; which is equivalent to finding α “ pα1, . . . , αmq

t suchthat

0 “M tpv ´Mαq, where M “

¨

˝

| |

u1 ¨ ¨ ¨ um| |

˛

‚PMn,mpF q.

Note: M is not square, but M tM PMmpF q is.

And since thecolumns of M are linearly independent, M tM is invertible. So

0 “M tpv ´Mαq ô M tMα “M tv ô α “ pM tMq´1M tv.

Then

u “ α1u1 ` ¨ ¨ ¨ ` αmum “Mα “MpM tMq´1M tv P U,

and uK “ v ´ u P UK. X

Let G grp of order n, F field of char - n, and consider V “ FGover R “ FG. Define x, y on FG by xg, hy “ δg,h and extendinglinearly.

Showing U ` UK “ V :Goal: for each v P V , find u P U and uK P U

K such thatv “ u` uK; which is equivalent to finding α “ pα1, . . . , αmq

t suchthat

0 “M tpv ´Mαq, where M “

¨

˝

| |

u1 ¨ ¨ ¨ um| |

˛

‚PMn,mpF q.

Note: M is not square, but M tM PMmpF q is. And since thecolumns of M are linearly independent, M tM is invertible.

So

0 “M tpv ´Mαq ô M tMα “M tv ô α “ pM tMq´1M tv.

Then

u “ α1u1 ` ¨ ¨ ¨ ` αmum “Mα “MpM tMq´1M tv P U,

and uK “ v ´ u P UK. X

Let G grp of order n, F field of char - n, and consider V “ FGover R “ FG. Define x, y on FG by xg, hy “ δg,h and extendinglinearly.

Showing U ` UK “ V :Goal: for each v P V , find u P U and uK P U

K such thatv “ u` uK; which is equivalent to finding α “ pα1, . . . , αmq

t suchthat

0 “M tpv ´Mαq, where M “

¨

˝

| |

u1 ¨ ¨ ¨ um| |

˛

‚PMn,mpF q.

Note: M is not square, but M tM PMmpF q is. And since thecolumns of M are linearly independent, M tM is invertible. So

0 “M tpv ´Mαq ô M tMα “M tv

ô α “ pM tMq´1M tv.

Then

u “ α1u1 ` ¨ ¨ ¨ ` αmum “Mα “MpM tMq´1M tv P U,

and uK “ v ´ u P UK. X

Let G grp of order n, F field of char - n, and consider V “ FGover R “ FG. Define x, y on FG by xg, hy “ δg,h and extendinglinearly.

Showing U ` UK “ V :Goal: for each v P V , find u P U and uK P U

K such thatv “ u` uK; which is equivalent to finding α “ pα1, . . . , αmq

t suchthat

0 “M tpv ´Mαq, where M “

¨

˝

| |

u1 ¨ ¨ ¨ um| |

˛

‚PMn,mpF q.

Note: M is not square, but M tM PMmpF q is. And since thecolumns of M are linearly independent, M tM is invertible. So

0 “M tpv ´Mαq ô M tMα “M tv ô α “ pM tMq´1M tv.

Then

u “ α1u1 ` ¨ ¨ ¨ ` αmum “Mα “MpM tMq´1M tv P U,

and uK “ v ´ u P UK. X

Let G grp of order n, F field of char - n, and consider V “ FGover R “ FG. Define x, y on FG by xg, hy “ δg,h and extendinglinearly.

Showing U ` UK “ V :Goal: for each v P V , find u P U and uK P U

K such thatv “ u` uK; which is equivalent to finding α “ pα1, . . . , αmq

t suchthat

0 “M tpv ´Mαq, where M “

¨

˝

| |

u1 ¨ ¨ ¨ um| |

˛

‚PMn,mpF q.

Note: M is not square, but M tM PMmpF q is. And since thecolumns of M are linearly independent, M tM is invertible. So

0 “M tpv ´Mαq ô M tMα “M tv ô α “ pM tMq´1M tv.

Then

u “ α1u1 ` ¨ ¨ ¨ ` αmum

“Mα “MpM tMq´1M tv P U,

and uK “ v ´ u P UK. X

Let G grp of order n, F field of char - n, and consider V “ FGover R “ FG. Define x, y on FG by xg, hy “ δg,h and extendinglinearly.

Showing U ` UK “ V :Goal: for each v P V , find u P U and uK P U

K such thatv “ u` uK; which is equivalent to finding α “ pα1, . . . , αmq

t suchthat

0 “M tpv ´Mαq, where M “

¨

˝

| |

u1 ¨ ¨ ¨ um| |

˛

‚PMn,mpF q.

Note: M is not square, but M tM PMmpF q is. And since thecolumns of M are linearly independent, M tM is invertible. So

0 “M tpv ´Mαq ô M tMα “M tv ô α “ pM tMq´1M tv.

Then

u “ α1u1 ` ¨ ¨ ¨ ` αmum “Mα

“MpM tMq´1M tv P U,

and uK “ v ´ u P UK. X

Let G grp of order n, F field of char - n, and consider V “ FGover R “ FG. Define x, y on FG by xg, hy “ δg,h and extendinglinearly.

Showing U ` UK “ V :Goal: for each v P V , find u P U and uK P U

K such thatv “ u` uK; which is equivalent to finding α “ pα1, . . . , αmq

t suchthat

0 “M tpv ´Mαq, where M “

¨

˝

| |

u1 ¨ ¨ ¨ um| |

˛

‚PMn,mpF q.

Note: M is not square, but M tM PMmpF q is. And since thecolumns of M are linearly independent, M tM is invertible. So

0 “M tpv ´Mαq ô M tMα “M tv ô α “ pM tMq´1M tv.

Then

u “ α1u1 ` ¨ ¨ ¨ ` αmum “Mα “MpM tMq´1M tv P U,

and uK “ v ´ u P UK. X

Result: Every v P V can be written as v “ u` uK, where

u “MpM tMq´1M tv P U and uK “ v ´ u P UK

(where M has column vectors corresponding to a basis of U).

Sanity check 1: When m “ 1, we’re back to M “ u and

MpM tMq´1M tv “ uputuq´1utv

“ upxu, uyq´1xu, vy “xv, uy

xu, uyu

as desired.Sanity check 2: If you started with an orthonormal basis of V(pairwise orthogonal and xvi, viy “ 1), and considered the spacespanned by the first m basis vectors, then

M “

1

1

. . .

0

m

. So M tM “ Im, and MpM tMq´1M t “ Im ‘ 0n´m.

Result: Every v P V can be written as v “ u` uK, where

u “MpM tMq´1M tv P U and uK “ v ´ u P UK

(where M has column vectors corresponding to a basis of U).

Sanity check 1: When m “ 1, we’re back to M “ u and

MpM tMq´1M tv “ uputuq´1utv “ upxu, uyq´1xu, vy

“xv, uy

xu, uyu

as desired.Sanity check 2: If you started with an orthonormal basis of V(pairwise orthogonal and xvi, viy “ 1), and considered the spacespanned by the first m basis vectors, then

M “

1

1

. . .

0

m

. So M tM “ Im, and MpM tMq´1M t “ Im ‘ 0n´m.

Result: Every v P V can be written as v “ u` uK, where

u “MpM tMq´1M tv P U and uK “ v ´ u P UK

(where M has column vectors corresponding to a basis of U).

Sanity check 1: When m “ 1, we’re back to M “ u and

MpM tMq´1M tv “ uputuq´1utv “ upxu, uyq´1xu, vy “xv, uy

xu, uyu

as desired.

Sanity check 2: If you started with an orthonormal basis of V(pairwise orthogonal and xvi, viy “ 1), and considered the spacespanned by the first m basis vectors, then

M “

1

1

. . .

0

m

. So M tM “ Im, and MpM tMq´1M t “ Im ‘ 0n´m.

Result: Every v P V can be written as v “ u` uK, where

u “MpM tMq´1M tv P U and uK “ v ´ u P UK

(where M has column vectors corresponding to a basis of U).

Sanity check 1: When m “ 1, we’re back to M “ u and

MpM tMq´1M tv “ uputuq´1utv “ upxu, uyq´1xu, vy “xv, uy

xu, uyu

as desired.Sanity check 2: If you started with an orthonormal basis of V(pairwise orthogonal and xvi, viy “ 1), and considered the spacespanned by the first m basis vectors, then

M “

1

1

. . .

0

m

.

So M tM “ Im, and MpM tMq´1M t “ Im ‘ 0n´m.

Result: Every v P V can be written as v “ u` uK, where

u “MpM tMq´1M tv P U and uK “ v ´ u P UK

(where M has column vectors corresponding to a basis of U).

Sanity check 1: When m “ 1, we’re back to M “ u and

MpM tMq´1M tv “ uputuq´1utv “ upxu, uyq´1xu, vy “xv, uy

xu, uyu

as desired.Sanity check 2: If you started with an orthonormal basis of V(pairwise orthogonal and xvi, viy “ 1), and considered the spacespanned by the first m basis vectors, then

M “

1

1

. . .

0

m

. So M tM “ Im,

and MpM tMq´1M t “ Im ‘ 0n´m.

Result: Every v P V can be written as v “ u` uK, where

u “MpM tMq´1M tv P U and uK “ v ´ u P UK

(where M has column vectors corresponding to a basis of U).

Sanity check 1: When m “ 1, we’re back to M “ u and

MpM tMq´1M tv “ uputuq´1utv “ upxu, uyq´1xu, vy “xv, uy

xu, uyu

as desired.Sanity check 2: If you started with an orthonormal basis of V(pairwise orthogonal and xvi, viy “ 1), and considered the spacespanned by the first m basis vectors, then

M “

1

1

. . .

0

m

. So M tM “ Im, and MpM tMq´1M t “ Im ‘ 0n´m.

In summary: G is a grp of order n, F is a field of char - n, and V “ FGis a module over R “ FG. Define x, y on FG by xg, hy “ δg,h andextending linearly. Then x, y is symmetric, bilinear, invariant, andnondegenerate; and for every submodule U Ď V , each v P V can bewritten as v “ u` uK, where

u “MpM tMq´1M tv P U and uK “ v ´ u P UK

(where M has column vectors corresponding to a basis of U). So

(1) UK is a submodule, (2) V “ U ` UK, and (3) U X UK “ 0.

Therefore V – U ‘ UK. ˝

Finally, note that

π “MpM tMq´1M t : V Ñ U

is a FG homomorphism:We have π|U “ idu, which is a homomorphism on U . And forv P V , writing v “ u` uK, we have

guK P UK (since x, y is G-invariant),

so that

πpgvq “ πpgu` guKq “ gu “ gπpuq.

In summary: G is a grp of order n, F is a field of char - n, and V “ FGis a module over R “ FG. Define x, y on FG by xg, hy “ δg,h andextending linearly. Then x, y is symmetric, bilinear, invariant, andnondegenerate; and for every submodule U Ď V , each v P V can bewritten as v “ u` uK, where

u “MpM tMq´1M tv P U and uK “ v ´ u P UK

(where M has column vectors corresponding to a basis of U). So

(1) UK is a submodule, (2) V “ U ` UK, and (3) U X UK “ 0.

Therefore V – U ‘ UK. ˝

Finally, note that

π “MpM tMq´1M t : V Ñ U

is a FG homomorphism:

We have π|U “ idu, which is a homomorphism on U . And forv P V , writing v “ u` uK, we have

guK P UK (since x, y is G-invariant),

so that

πpgvq “ πpgu` guKq “ gu “ gπpuq.

In summary: G is a grp of order n, F is a field of char - n, and V “ FGis a module over R “ FG. Define x, y on FG by xg, hy “ δg,h andextending linearly. Then x, y is symmetric, bilinear, invariant, andnondegenerate; and for every submodule U Ď V , each v P V can bewritten as v “ u` uK, where

u “MpM tMq´1M tv P U and uK “ v ´ u P UK

(where M has column vectors corresponding to a basis of U). So

(1) UK is a submodule, (2) V “ U ` UK, and (3) U X UK “ 0.

Therefore V – U ‘ UK. ˝

Finally, note that

π “MpM tMq´1M t : V Ñ U

is a FG homomorphism:We have π|U “ idu, which is a homomorphism on U .

And forv P V , writing v “ u` uK, we have

guK P UK (since x, y is G-invariant),

so that

πpgvq “ πpgu` guKq “ gu “ gπpuq.

In summary: G is a grp of order n, F is a field of char - n, and V “ FGis a module over R “ FG. Define x, y on FG by xg, hy “ δg,h andextending linearly. Then x, y is symmetric, bilinear, invariant, andnondegenerate; and for every submodule U Ď V , each v P V can bewritten as v “ u` uK, where

u “MpM tMq´1M tv P U and uK “ v ´ u P UK

(where M has column vectors corresponding to a basis of U). So

(1) UK is a submodule, (2) V “ U ` UK, and (3) U X UK “ 0.

Therefore V – U ‘ UK. ˝

Finally, note that

π “MpM tMq´1M t : V Ñ U

is a FG homomorphism:We have π|U “ idu, which is a homomorphism on U . And forv P V , writing v “ u` uK, we have

guK P UK (since x, y is G-invariant),

so that

πpgvq “ πpgu` guKq “ gu “ gπpuq.

In summary: G is a grp of order n, F is a field of char - n, and V “ FGis a module over R “ FG. Define x, y on FG by xg, hy “ δg,h andextending linearly. Then x, y is symmetric, bilinear, invariant, andnondegenerate; and for every submodule U Ď V , each v P V can bewritten as v “ u` uK, where

u “MpM tMq´1M tv P U and uK “ v ´ u P UK

(where M has column vectors corresponding to a basis of U). So

(1) UK is a submodule, (2) V “ U ` UK, and (3) U X UK “ 0.

Therefore V – U ‘ UK. ˝

Finally, note that

π “MpM tMq´1M t : V Ñ U

is a FG homomorphism:We have π|U “ idu, which is a homomorphism on U . And forv P V , writing v “ u` uK, we have

guK P UK (since x, y is G-invariant),

so that

πpgvq “ πpgu` guKq “ gu “ gπpuq.

What made this proof work?

From the perspective of the ring:FG is an algebra over a field, so fin-dim’l modules are all vectorspaces (isomorphic to F d for d “ dimpV q).

Using the standard dotproduct on F d, the projection π produces an orthogonalcomplement UK to any subspace U Ď V .So V – U ‘ UK as vector spaces.

What’s missing: under what circumstances is UK a submodule?

In our example, invariance was a result of

Mpρpgqq´1 “Mpρpgqqt for all g P G.

Not always true! In particular, it requires tools specific to themodule, not the ring. And, in fact, the complementary directsummand is often not the orthogonal complement.

Further, if you use an inner product other than dot product, thenyou might get U ` UK Ĺ V .

What made this proof work?

From the perspective of the ring:FG is an algebra over a field, so fin-dim’l modules are all vectorspaces (isomorphic to F d for d “ dimpV q). Using the standard dotproduct on F d, the projection π produces an orthogonalcomplement UK to any subspace U Ď V .So V – U ‘ UK as vector spaces.

What’s missing: under what circumstances is UK a submodule?

In our example, invariance was a result of

Mpρpgqq´1 “Mpρpgqqt for all g P G.

Not always true! In particular, it requires tools specific to themodule, not the ring. And, in fact, the complementary directsummand is often not the orthogonal complement.

Further, if you use an inner product other than dot product, thenyou might get U ` UK Ĺ V .

What made this proof work?

From the perspective of the ring:FG is an algebra over a field, so fin-dim’l modules are all vectorspaces (isomorphic to F d for d “ dimpV q). Using the standard dotproduct on F d, the projection π produces an orthogonalcomplement UK to any subspace U Ď V .So V – U ‘ UK as vector spaces.

What’s missing: under what circumstances is UK a submodule?

In our example, invariance was a result of

Mpρpgqq´1 “Mpρpgqqt for all g P G.

Not always true! In particular, it requires tools specific to themodule, not the ring. And, in fact, the complementary directsummand is often not the orthogonal complement.

Further, if you use an inner product other than dot product, thenyou might get U ` UK Ĺ V .

What made this proof work?

From the perspective of the ring:FG is an algebra over a field, so fin-dim’l modules are all vectorspaces (isomorphic to F d for d “ dimpV q). Using the standard dotproduct on F d, the projection π produces an orthogonalcomplement UK to any subspace U Ď V .So V – U ‘ UK as vector spaces.

What’s missing: under what circumstances is UK a submodule?

In our example, invariance was a result of

Mpρpgqq´1 “Mpρpgqqt for all g P G.

Not always true! In particular, it requires tools specific to themodule, not the ring. And, in fact, the complementary directsummand is often not the orthogonal complement.

Further, if you use an inner product other than dot product, thenyou might get U ` UK Ĺ V .

What made this proof work?

From the perspective of the ring:FG is an algebra over a field, so fin-dim’l modules are all vectorspaces (isomorphic to F d for d “ dimpV q). Using the standard dotproduct on F d, the projection π produces an orthogonalcomplement UK to any subspace U Ď V .So V – U ‘ UK as vector spaces.

What’s missing: under what circumstances is UK a submodule?

In our example, invariance was a result of

Mpρpgqq´1 “Mpρpgqqt for all g P G.

Not always true! In particular, it requires tools specific to themodule, not the ring. And, in fact, the complementary directsummand is often not the orthogonal complement.

Further, if you use an inner product other than dot product, thenyou might get U ` UK Ĺ V .

Example

Let R “ V “ CS2. Fix basis v1 “ 1 and v2 “ p12q, and identify Vwith C2 and EndpV q with M2pCq.

Consider the bilinear form

x, y “ x, yJ where J “

ˆ

1 00 ´1

˙

.

Nondegenerate X Symmetric X

Not invariant:

xv1, v1y “ p1, 0q

ˆ

1 00 ´1

˙ˆ

10

˙

“ p1, 0q

ˆ

10

˙

“ 1

xv2, v2y “ p0, 1q

ˆ

1 00 ´1

˙ˆ

01

˙

“ p0, 1q

ˆ

0´1

˙

“ ´1,

So

xp12qv1, p12qv1y “ xv2, v2y ‰ xv1, v1y.

Example

Let R “ V “ CS2. Fix basis v1 “ 1 and v2 “ p12q, and identify Vwith C2 and EndpV q with M2pCq.

Consider the bilinear form

x, y “ x, yJ where J “

ˆ

1 00 ´1

˙

.

Nondegenerate X Symmetric X

Not invariant:

xv1, v1y “ p1, 0q

ˆ

1 00 ´1

˙ˆ

10

˙

“ p1, 0q

ˆ

10

˙

“ 1

xv2, v2y “ p0, 1q

ˆ

1 00 ´1

˙ˆ

01

˙

“ p0, 1q

ˆ

0´1

˙

“ ´1,

So

xp12qv1, p12qv1y “ xv2, v2y ‰ xv1, v1y.

Example

Let R “ V “ CS2. Fix basis v1 “ 1 and v2 “ p12q, and identify Vwith C2 and EndpV q with M2pCq.

Consider the bilinear form

x, y “ x, yJ where J “

ˆ

1 00 ´1

˙

.

Nondegenerate X Symmetric X

Not invariant:

xv1, v1y “ p1, 0q

ˆ

1 00 ´1

˙ˆ

10

˙

“ p1, 0q

ˆ

10

˙

“ 1

xv2, v2y “ p0, 1q

ˆ

1 00 ´1

˙ˆ

01

˙

“ p0, 1q

ˆ

0´1

˙

“ ´1,

So

xp12qv1, p12qv1y “ xv2, v2y ‰ xv1, v1y.

Example

Let R “ V “ CS2. Fix basis v1 “ 1 and v2 “ p12q, and identify Vwith C2 and EndpV q with M2pCq.

Consider the bilinear form

x, y “ x, yJ where J “

ˆ

1 00 ´1

˙

.

Nondegenerate X Symmetric X

Not invariant:

xv1, v1y “ p1, 0q

ˆ

1 00 ´1

˙ˆ

10

˙

“ p1, 0q

ˆ

10

˙

“ 1

xv2, v2y “ p0, 1q

ˆ

1 00 ´1

˙ˆ

01

˙

“ p0, 1q

ˆ

0´1

˙

“ ´1,

So

xp12qv1, p12qv1y “ xv2, v2y ‰ xv1, v1y.

Example

Let R “ V “ CS2. Fix basis v1 “ 1 and v2 “ p12q, and identify Vwith C2 and EndpV q with M2pCq.

Consider the bilinear form

x, y “ x, yJ where J “

ˆ

1 00 ´1

˙

.

Nondegenerate X Symmetric X

Not invariant:

xv1, v1y “ p1, 0q

ˆ

1 00 ´1

˙ˆ

10

˙

“ p1, 0q

ˆ

10

˙

“ 1

xv2, v2y “ p0, 1q

ˆ

1 00 ´1

˙ˆ

01

˙

“ p0, 1q

ˆ

0´1

˙

“ ´1,

So

xp12qv1, p12qv1y “ xv2, v2y ‰ xv1, v1y.

Example

Let R “ V “ CS2. Fix basis v1 “ 1 and v2 “ p12q, and identify Vwith C2 and EndpV q with M2pCq.

Consider the bilinear form

x, y “ x, yJ where J “

ˆ

1 00 ´1

˙

.

Nondegenerate X Symmetric X Not invariant ˆ

Now try to decompose CS2. . .We have the submodule T “ Ce`, where e` “ v1 ` v2.Further,

T K “ tv P CS2 | xv1 ` v2, vy “ 0u

So v “ αv1 ` αv2 P T K whenever

0 “ xv1 ` v2, vy “ pα1, α2q

ˆ

1 00 ´1

˙ˆ

11

˙

“ α1 ´ α2.

But that implies that T K “ T , so that T ` T K “ T ‰ V .(Compare to the known decomposition CS2 – T ‘ S, where S “ Ce´.)

Example

Let R “ V “ CS2. Fix basis v1 “ 1 and v2 “ p12q, and identify Vwith C2 and EndpV q with M2pCq.

Consider the bilinear form

x, y “ x, yJ where J “

ˆ

1 00 ´1

˙

.

Nondegenerate X Symmetric X Not invariant ˆ

Now try to decompose CS2. . .We have the submodule T “ Ce`, where e` “ v1 ` v2.

Further,

T K “ tv P CS2 | xv1 ` v2, vy “ 0u

So v “ αv1 ` αv2 P T K whenever

0 “ xv1 ` v2, vy “ pα1, α2q

ˆ

1 00 ´1

˙ˆ

11

˙

“ α1 ´ α2.

But that implies that T K “ T , so that T ` T K “ T ‰ V .(Compare to the known decomposition CS2 – T ‘ S, where S “ Ce´.)

Example

Let R “ V “ CS2. Fix basis v1 “ 1 and v2 “ p12q, and identify Vwith C2 and EndpV q with M2pCq.

Consider the bilinear form

x, y “ x, yJ where J “

ˆ

1 00 ´1

˙

.

Nondegenerate X Symmetric X Not invariant ˆ

Now try to decompose CS2. . .We have the submodule T “ Ce`, where e` “ v1 ` v2.Further,

T K “ tv P CS2 | xv1 ` v2, vy “ 0u

So v “ αv1 ` αv2 P T K whenever

0 “ xv1 ` v2, vy “ pα1, α2q

ˆ

1 00 ´1

˙ˆ

11

˙

“ α1 ´ α2.

But that implies that T K “ T , so that T ` T K “ T ‰ V .(Compare to the known decomposition CS2 – T ‘ S, where S “ Ce´.)

Example

Let R “ V “ CS2. Fix basis v1 “ 1 and v2 “ p12q, and identify Vwith C2 and EndpV q with M2pCq.

Consider the bilinear form

x, y “ x, yJ where J “

ˆ

1 00 ´1

˙

.

Nondegenerate X Symmetric X Not invariant ˆ

Now try to decompose CS2. . .We have the submodule T “ Ce`, where e` “ v1 ` v2.Further,

T K “ tv P CS2 | xv1 ` v2, vy “ 0u

So v “ αv1 ` αv2 P T K whenever

0 “ xv1 ` v2, vy “ pα1, α2q

ˆ

1 00 ´1

˙ˆ

11

˙

“ α1 ´ α2.

But that implies that T K “ T , so that T ` T K “ T ‰ V .(Compare to the known decomposition CS2 – T ‘ S, where S “ Ce´.)

Example

Let R “ V “ CS2. Fix basis v1 “ 1 and v2 “ p12q, and identify Vwith C2 and EndpV q with M2pCq.

Consider the bilinear form

x, y “ x, yJ where J “

ˆ

1 00 ´1

˙

.

Nondegenerate X Symmetric X Not invariant ˆ

Now try to decompose CS2. . .We have the submodule T “ Ce`, where e` “ v1 ` v2.Further,

T K “ tv P CS2 | xv1 ` v2, vy “ 0u

So v “ αv1 ` αv2 P T K whenever

0 “ xv1 ` v2, vy “ pα1, α2q

ˆ

1 00 ´1

˙ˆ

11

˙

“ α1 ´ α2.

But that implies that T K “ T , so that T ` T K “ T ‰ V .(Compare to the known decomposition CS2 – T ‘ S, where S “ Ce´.)

Example

Let R “ V “ CS2. Fix basis v1 “ 1 and v2 “ p12q, and identify Vwith C2 and EndpV q with M2pCq.

Consider the bilinear form

x, y “ x, yJ where J “

ˆ

1 00 ´1

˙

.

Nondegenerate X Symmetric X Not invariant ˆ

Now try to decompose CS2. . .We have the submodule T “ Ce`, where e` “ v1 ` v2.Further,

T K “ tv P CS2 | xv1 ` v2, vy “ 0u

So v “ αv1 ` αv2 P T K whenever

0 “ xv1 ` v2, vy “ pα1, α2q

ˆ

1 00 ´1

˙ˆ

11

˙

“ α1 ´ α2.

But that implies that T K “ T

, so that T ` T K “ T ‰ V .(Compare to the known decomposition CS2 – T ‘ S, where S “ Ce´.)

Example

Let R “ V “ CS2. Fix basis v1 “ 1 and v2 “ p12q, and identify Vwith C2 and EndpV q with M2pCq.

Consider the bilinear form

x, y “ x, yJ where J “

ˆ

1 00 ´1

˙

.

Nondegenerate X Symmetric X Not invariant ˆ

Now try to decompose CS2. . .We have the submodule T “ Ce`, where e` “ v1 ` v2.Further,

T K “ tv P CS2 | xv1 ` v2, vy “ 0u

So v “ αv1 ` αv2 P T K whenever

0 “ xv1 ` v2, vy “ pα1, α2q

ˆ

1 00 ´1

˙ˆ

11

˙

“ α1 ´ α2.

But that implies that T K “ T , so that T ` T K “ T ‰ V .(Compare to the known decomposition CS2 – T ‘ S, where S “ Ce´.)

Example

Again, let R “ V “ CS2. Fix basis v1 “ 1 and v2 “ p12q, andidentify V with C2 and EndpV q with M2pCq.Now consider

x, y “ x, yJ where J “

ˆ

1 00 2

˙

Nondegenerate X Symmetric X

Not invariant:

xv1, v1y “ p1, 0q

ˆ

1 00 2

˙ˆ

10

˙

“ 1

xv2, v2y “ p0, 1q

ˆ

1 00 2

˙ˆ

01

˙

“ 2 ‰ 1,

so that x, y is not S2-invariant.

Example

Again, let R “ V “ CS2. Fix basis v1 “ 1 and v2 “ p12q, andidentify V with C2 and EndpV q with M2pCq.Now consider

x, y “ x, yJ where J “

ˆ

1 00 2

˙

Nondegenerate X Symmetric X

Not invariant:

xv1, v1y “ p1, 0q

ˆ

1 00 2

˙ˆ

10

˙

“ 1

xv2, v2y “ p0, 1q

ˆ

1 00 2

˙ˆ

01

˙

“ 2 ‰ 1,

so that x, y is not S2-invariant.

Example: Again, let R “ V “ CS2. Fix basis v1 “ 1 andv2 “ p12q, and identify V with C2 and EndpV q with M2pCq.Now consider

x, y “ x, yJ where J “

ˆ

1 00 2

˙

Nondegenerate X Symmetric X Invariant ˆ

Again, let T “ Ce` and consider T K: v “ αv1 ` αv2 P T Kwhenever

0 “ xv1 ` v2, vy “ pα1, α2q

ˆ

1 00 2

˙ˆ

11

˙

“ α1 ` 2α2.

So T K “ Cp2v1 ´ v2q.

This time, we do have T X T K “ 0 and T ` T K “ V , so thatV – T ‘ T K as vector spaces.

However, this is still the wrong decomposition of V as modules,since T K is not closed under the S2 action:

p12qp2v1 ´ v2q “ ´v1 ` 2v2 R T K. So V fl T ‘ T K.

Example: Again, let R “ V “ CS2. Fix basis v1 “ 1 andv2 “ p12q, and identify V with C2 and EndpV q with M2pCq.Now consider

x, y “ x, yJ where J “

ˆ

1 00 2

˙

Nondegenerate X Symmetric X Invariant ˆ

Again, let T “ Ce` and consider T K:

v “ αv1 ` αv2 P T Kwhenever

0 “ xv1 ` v2, vy “ pα1, α2q

ˆ

1 00 2

˙ˆ

11

˙

“ α1 ` 2α2.

So T K “ Cp2v1 ´ v2q.

This time, we do have T X T K “ 0 and T ` T K “ V , so thatV – T ‘ T K as vector spaces.

However, this is still the wrong decomposition of V as modules,since T K is not closed under the S2 action:

p12qp2v1 ´ v2q “ ´v1 ` 2v2 R T K. So V fl T ‘ T K.

Example: Again, let R “ V “ CS2. Fix basis v1 “ 1 andv2 “ p12q, and identify V with C2 and EndpV q with M2pCq.Now consider

x, y “ x, yJ where J “

ˆ

1 00 2

˙

Nondegenerate X Symmetric X Invariant ˆ

Again, let T “ Ce` and consider T K: v “ αv1 ` αv2 P T Kwhenever

0 “ xv1 ` v2, vy “ pα1, α2q

ˆ

1 00 2

˙ˆ

11

˙

“ α1 ` 2α2.

So T K “ Cp2v1 ´ v2q.

This time, we do have T X T K “ 0 and T ` T K “ V , so thatV – T ‘ T K as vector spaces.

However, this is still the wrong decomposition of V as modules,since T K is not closed under the S2 action:

p12qp2v1 ´ v2q “ ´v1 ` 2v2 R T K. So V fl T ‘ T K.

Example: Again, let R “ V “ CS2. Fix basis v1 “ 1 andv2 “ p12q, and identify V with C2 and EndpV q with M2pCq.Now consider

x, y “ x, yJ where J “

ˆ

1 00 2

˙

Nondegenerate X Symmetric X Invariant ˆ

Again, let T “ Ce` and consider T K: v “ αv1 ` αv2 P T Kwhenever

0 “ xv1 ` v2, vy “ pα1, α2q

ˆ

1 00 2

˙ˆ

11

˙

“ α1 ` 2α2.

So T K “ Cp2v1 ´ v2q.

This time, we do have T X T K “ 0 and T ` T K “ V , so thatV – T ‘ T K as vector spaces.

However, this is still the wrong decomposition of V as modules,since T K is not closed under the S2 action:

p12qp2v1 ´ v2q “ ´v1 ` 2v2 R T K. So V fl T ‘ T K.

Example: Again, let R “ V “ CS2. Fix basis v1 “ 1 andv2 “ p12q, and identify V with C2 and EndpV q with M2pCq.Now consider

x, y “ x, yJ where J “

ˆ

1 00 2

˙

Nondegenerate X Symmetric X Invariant ˆ

Again, let T “ Ce` and consider T K: v “ αv1 ` αv2 P T Kwhenever

0 “ xv1 ` v2, vy “ pα1, α2q

ˆ

1 00 2

˙ˆ

11

˙

“ α1 ` 2α2.

So T K “ Cp2v1 ´ v2q.

This time, we do have T X T K “ 0 and T ` T K “ V , so thatV – T ‘ T K as vector spaces.

However, this is still the wrong decomposition of V as modules,since T K is not closed under the S2 action:

p12qp2v1 ´ v2q “ ´v1 ` 2v2 R T K. So V fl T ‘ T K.

Example: Again, let R “ V “ CS2. Fix basis v1 “ 1 andv2 “ p12q, and identify V with C2 and EndpV q with M2pCq.Now consider

x, y “ x, yJ where J “

ˆ

1 00 2

˙

Nondegenerate X Symmetric X Invariant ˆ

Again, let T “ Ce` and consider T K: v “ αv1 ` αv2 P T Kwhenever

0 “ xv1 ` v2, vy “ pα1, α2q

ˆ

1 00 2

˙ˆ

11

˙

“ α1 ` 2α2.

So T K “ Cp2v1 ´ v2q.

This time, we do have T X T K “ 0 and T ` T K “ V , so thatV – T ‘ T K as vector spaces.

However, this is still the wrong decomposition of V as modules,since T K is not closed under the S2 action:

p12qp2v1 ´ v2q “ ´v1 ` 2v2 R T K.

So V fl T ‘ T K.

Example: Again, let R “ V “ CS2. Fix basis v1 “ 1 andv2 “ p12q, and identify V with C2 and EndpV q with M2pCq.Now consider

x, y “ x, yJ where J “

ˆ

1 00 2

˙

Nondegenerate X Symmetric X Invariant ˆ

Again, let T “ Ce` and consider T K: v “ αv1 ` αv2 P T Kwhenever

0 “ xv1 ` v2, vy “ pα1, α2q

ˆ

1 00 2

˙ˆ

11

˙

“ α1 ` 2α2.

So T K “ Cp2v1 ´ v2q.

This time, we do have T X T K “ 0 and T ` T K “ V , so thatV – T ‘ T K as vector spaces.

However, this is still the wrong decomposition of V as modules,since T K is not closed under the S2 action:

p12qp2v1 ´ v2q “ ´v1 ` 2v2 R T K. So V fl T ‘ T K.