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:VT(V) is a linear transformation, and
  2. if E is an algebra and f:VE is a linear transformation then there exists a unique algebra homomorphism g:T(V)E such that f=gφ.
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:VS(V) is a linear transformation such that if   v1,v2V   then   ι(v1)ι(v2) = ι(v2)ι(v1),  and
  2. if E is an algebra and f:VE is a linear transformation such that if   v1,v2V   then   f(v1)f(v2) = f(v2)f(v1),  then there exists a unique algebra homomorphism g:S(V)E such that f=gι.
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:VΛ(V) is a linear transformation such that if   vV   then   ι(v)2 = 0,  and
  2. if E is an algebra and f:VE is a linear transformation such that if   vV   then   f(v)2 = 0,  then there exists a unique homomorphism of algebras g:Λ(V)E such that f=gι.
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 Vk = VVV k   factors by if b1...bn   is a basis of   V then {bi1bik  |  1i1,...,ikn}   is a basis of   Vk and if u:VW is a linear transformation then Tku: Vk Wk is given by (Tku) (bi1bik) =u(bi1)u(bik) with bi1 (c1bi+c2bj) lth factor bik = c1bi1 bi lth factor bik + c2bi1 bj lth factor bik for c1,c2𝔽 and l{1,...,k}.

Then T(V) = k0 Vk with V0 = 𝔽, (bi1bik) (bj1bjl) = bi1bikbj1bjl and ι: V k0 Vk 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 {xi1xik  |  1i1i2ikn}   is a basis of   Sk(V) and if u:VW is a linear transformation then Sku: Sk(V) Sk(W) is given by (Sku) (xi1xik) = (uxi1) (uxi2) (uxik) with xi1 (c1xi+c2xj) lth factor xik = c1xi1 xi lth factor xik + c2xi1 xj lth factor xik for c1,c2𝔽 and l{1,...,k} and xi1xik = xiσ(1)xiσ(k) for   σSk.

Then S(V) = k0 Sk(V) with S0(V) = 𝔽, (xi1xik) (xj1xjl) = xi1xikxj1xjl 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 {ei1eik  |  1i1<i2<<ikn}   is a basis of   ΛkV and if u:VW is a linear transformation then Λku: ΛkV ΛkW is given by (Λku) (ei1eik) = (uei1) (ueik) with ei1 (c1ei+c2ej) lth factor eik = c1ei1 ei lth factor eik + c2ei1 ej lth factor eik for c1,c2𝔽 and l{1,...,k} and ei1eik = det(σ) eiσ(1)eiσ(k),   for   σSk.

Homework problems

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

HW: Identifying a matrix AMn×m(𝔽) with a linear transformation u:𝔽m𝔽n 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 σSk. 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 AMn(𝔽) be identified with a linear transformation u:𝔽n𝔽n. Compute ΛnA, the matrix corresponding to the linear transformation   Λnu, in terms of the matrix entries of A.

HW: Let AMn(𝔽) be identified with a linear transformation u:𝔽n𝔽n. 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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