Tensor, symmetric and exterior algebras

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last update: 17 March 2012

Definition of the algebras

Let V be a vector space.

The tensor algebra (T(V),φ) is the pair with

  1. T(V) is an algebra and φ:V→T(V) is a linear transformation, and
  2. if E is an algebra and f:V→E is a linear transformation then there exists a unique algebra homomorphism g:T(V)→E such that f=g∘φ.
V T(V) E φ f g

The symmetric algebra (S(V),ι) is the pair with

  1. S(V) is an algebra and ι:V→S(V) is a linear transformation such that if   v1,v2∈V   then   ι(v1)ι(v2) = ι(v2)ι(v1),  and
  2. if E is an algebra and f:V→E is a linear transformation such that if   v1,v2∈V   then   f(v1)f(v2) = f(v2)f(v1),  then there exists a unique algebra homomorphism g:S(V)→E such that f=g∘ι.
V S(V) E ι f g

The exterior algebra is the pair (Λ(V),ι) where

  1. Λ(V) is an algebra and ι:V→Λ(V) is a linear transformation such that if   v∈V   then   ι(v)2 = 0,  and
  2. if E is an algebra and f:V→E is a linear transformation such that if   v∈V   then   f(v)2 = 0,  then there exists a unique homomorphism of algebras g:Λ(V)→E such that f=g∘ι.
V Λ(V) E ι f g

The functors Tk, Sk,  and   Λk

The functor Tk

Define a functor Tk: { vector spaces } → { vector spaces } V ↦ V⊗k = V⊗V⊗⋯⊗V ⏟ k   factors by if b1...bn   is a basis of   V then {bi1⊗⋯⊗bik  |  1≤i1,...,ik≤n}   is a basis of   V⊗k and if u:V→W is a linear transformation then Tku: V⊗k→ W⊗k is given by (Tku) (bi1⊗⋯⊗bik) =u(bi1)⊗⋯⊗u(bik) with bi1⊗⋯⊗ (c1bi+c2bj) lth factor ⊗⋯⊗bik = c1bi1⊗⋯ ⊗bi⊗ lth factor ⋯⊗bik + c2bi1⊗⋯ ⊗bj⊗ lth factor ⋯⊗bik for c1,c2∈𝔽 and l∈{1,...,k}.

Then T(V) = ⨁k∈ℤ≥0 V⊗k with V⊗0 = 𝔽, (bi1⊗⋯⊗bik) (bj1⊗⋯⊗bjl) = bi1⊗⋯⊗bik⊗bj1⊗⋯⊗bjl and ι: V → ⊕k∈ℤ≥0 V⊗k bi ↦ bi.

The functor Sk

Define a functor Sk: { vector spaces } → { vector spaces } V ↦ Sk(V) by if x1...xn   is a basis of   V then {xi1⋯xik  |  1≤i1≤i2≤⋯≤ik≤n}   is a basis of   Sk(V) and if u:V→W is a linear transformation then Sku: Sk(V) → Sk(W) is given by (Sku) (xi1⋯xik) = (uxi1) (uxi2) ⋯ (uxik) with xi1⋯ (c1xi+c2xj) lth factor ⋯xik = c1xi1 ⋯xi⋯ lth factor xik + c2xi1 ⋯xj⋯ lth factor xik for c1,c2∈𝔽 and l∈{1,...,k} and xi1⋯xik = xiσ(1)⋯xiσ(k) for   σ∈Sk.

Then S(V) = ⨁k∈ℤ≥0 Sk(V) with S0(V) = 𝔽, (xi1⋯xik) (xj1⋯xjl) = xi1⋯xikxj1⋯xjl and ι: V → S(V) xi ↦ xi.

The functor Λk

Define a functor Λk: { vector spaces } → { vector spaces } V ↦ ΛkV by if e1...en   is a basis of   V then {ei1∧⋯∧eik  |  1≤i1<i2<⋯<ik≤n}   is a basis of   ΛkV and if u:V→W is a linear transformation then Λku: ΛkV→ ΛkW is given by (Λku) (ei1∧⋯∧eik) = (uei1) ∧⋯∧ (ueik) with ei1∧⋯∧ (c1ei+c2ej) lth factor ∧⋯∧eik = c1ei1∧⋯ ∧ei∧ lth factor ⋯∧eik + c2ei1∧⋯ ∧ej∧ lth factor ⋯∧eik for c1,c2∈𝔽 and l∈{1,...,k} and ei1∧⋯∧eik = det(σ) eiσ(1)∧⋯∧eiσ(k),   for   σ∈Sk.

Homework problems

HW: If dim(V)=n compute dim(Tk(V)), dim(Sk(V)) and dim(Λk(V)).

HW: Identifying a matrix A∈Mn×m(𝔽) with a linear transformation u:𝔽m→𝔽n compute the matrix TkA corresponding to Tku SkA corresponding to Sku ΛkA corresponding to Λku. First do this for k=1 and 2.

HW: Let σ∈Sk. Show that det(σ) = (-1)l(σ) where l(σ) = Card(R(σ)) and R(σ) = {(i,j)  |  i,j∈{1,2,...,k},  i<j   and   σ(i)>σ(j)}. Interpret l(σ)   as the 'number of crossings in   σ'.

HW: Let A∈Mn(𝔽) be identified with a linear transformation u:𝔽n→𝔽n. Compute ΛnA, the matrix corresponding to the linear transformation   Λnu, in terms of the matrix entries of A.

HW: Let A∈Mn(𝔽) be identified with a linear transformation u:𝔽n→𝔽n. Compute ΛkA, the matrix corresponding to the linear transformation   Λku, in terms of the matrix entries of A.

HW: Show that if dim(V)=n then dim(T(V)) = 1 1-n .

Notes and References

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References

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