The tensor algebra

The tensor algebra - definition

Suppose that $R$ is a commutative ring with unity. Let $M$ be an $R$ module.

If $T^j(M)=\stackrel{j}{\overbrace{M{\otimes }_R\cdots {\otimes }_{R}M}}$

then we can define a (non-commutative) product map

$$T^j(M)\times T^k(M)\to T^{j+k}(M)$$

by defining

$$(a_1\otimes \cdots \otimes a_j)(b_1\otimes \cdots \otimes b_k)=(a_1\otimes \cdots \otimes a_j\otimes b_1\otimes \cdots \otimes b_k)$$

This allows us to view the infinite direct sum

$$T(M)=R\oplus M\oplus T^2(M)\oplus T^3(M)\cdots$$

as a (non-commutative) $R$-algebra. This ring is called the tensor algebra of $M$.

Universal property

If $A$ is any $R$-algebra and $M$ is any $R$ module, then given a map $f:M\to A$ there is a unique map $T(M)\to A$ which, when restricted to $M$ gives $f$.

Graded rings

The tensor algebra $T(M)$ is an example of a graded ring. A general graded ring is any ring $S$ that is a direct sum of additive subgroups $S_i$ where $S_i{S}_j\subset S_{i+j}$. Each $S_j$ is called the subset of homogeneous elements of degree $j$.

A graded ideal $I$ is an ideal of $S$ that is the direct sum of its homogeneous components $I\cap S_j$.

A map $f:S\to T$ between graded rings is a homomorphism of graded rings if $f(S_j)\subset T_j$ for every $j$.

Example: The polynomial ring $F[x_1,\dots ,x_n]$ is graded where the homogeneous components are spanned by the monomials of a given degree. The ideal generated by $x_1,\dots ,x_n$ is graded because any element of this ideal – that is, any polynomiial with zero constant term – can be written as a sum of monomials of a fixed degree in a unique way.

The ring $R=\mathbb{Z}[x]$ is graded by degree, and the ideal generated by $J=(1+x)$ is not graded because $1$ and $x$ are not in $J$.

An ideal generated by homogeneous elements in a graded ring is graded.

Quotients of graded rings

If $S$ is a graded ring and $I$ is a graded ideal, then $S/I$ is graded with homogeneous components $S_j/I_j$.

The symmetric algebra

The symmetric algebra - definition

A bilinear map $f:M\times M\to L$ is symmetric if $f(m_1,m_2)=f(m_2,m_1)$ for all pairs $(m_1,m_2)$. The symmetric tensor product $S^2(M)$ has the universal property that any symmetric bilinar map $f:M\times M\to L$ “factors through” $S^2(M)$ in a unique way. $S^2(M)$ is constructed from $M\otimes M$ by imposing the relation $m\otimes n-n\otimes m=0$ for all pairs $(m,n)$.

The symmetric algebra $S(M)$ is obtained from the tensor algebra by modding out by the ideal $C(M)$ generated by all $m\otimes n-n\otimes m$ in $T(M)$. The map

$$T(M)\to S(M)$$

is a map of graded rings since $C(M)$ is a graded ideal, so $S(M)$ is a graded ring. $S^0(M)=R$ and $S^1(M)=M$.

More on the symmetric algebra

In $S(M)$, the degree $k$ terms are spanned by tensors $m_1\otimes \cdots \otimes m_k$ modulo the relation which allows you to freely permute the terms.

Any $k$-symmetric multilinear map $M\times \cdots \times M\to L$ factors through $S^k(M)$ in a unique way.

If $A$ is any commutative $R$ algebra, and $f:M\to A$ is a module map, then there is a unique $R$-algebra map $S(M)\to A$ which restricts to $f$ on $M$.

If $R$ is a field and $V$ is a vector space of dimension $k$ then $S(V)$ is the (graded) polynomial ring in $k$ variables over $R$. If $R$ is a commutative ring and $V$ is a free module of rank $k$ then $S(V)$ is the (graded) polynomial ring over $R$ in $k$ variables.

The exterior algebra

Alternating maps and the wedge product

A bilinear map $f:M\times M\to L$ is alternating if $f(m,m)=0$ for all $m\in M$. The exterior product (or wedge product) $M\wedge M$ has the property that there is a map $M\times M\to M\wedge M$ such that if $f:M\times M\to L$ is an alternating map, then there is a unique map $M\wedge M\to L$ making the usual triangle commute.

The map $M\otimes M\to M\wedge M$ is written $(m_1,m_2)\to m_1\wedge m_2$ and $M\wedge M$ is spanned by all $m_1\wedge m_2$. We have $m_1\wedge m_2=-m_2\wedge m_1$ for elementary tensors where $m_1,m_2\in M$.

$M\wedge M$ is the quotient of $M\otimes M$ by the elementary tensors $m\otimes m$.

The exterior algebra

The quotient of the tensor algebra by the ideal generated by all tensors of the form $m\otimes m$ is called the exterior algebra, written $\bigwedge M$. It is graded so that that the degree $k$ terms consist of linear combinations of the “wedge” of $k$ elements of $M$.

If $A$ is an $R$-algebra such that $a^2=0$ for all $a\in A$, and $f:M\to A$ is a module homomorphism, then there is a unique map $\bigwedge M\to A$ which restricts to $f$.

If $V$ is a finite dimensional vector space over a field $F$, then $\bigwedge V$ is finite dimensional. This holds for any free, finite rank module over a ring $R$.

If $R=\mathbb{Z}[x,y]$, if $M=R$ then $\bigwedge M$ is spanned by $1\wedge 1=0$ so $\bigwedge M=0$. If $I=(x,y)$, then there is an alternating map from $I$ to $\mathbb{Z}$ defined by

$$f(ax+by,cx+dy)=ad-bc(\mathrm{mod}(x,y))$$

so $x\wedge y$ is not zero.

The cross product in ${\mathbb{R}}^3$.

The coefficients of

$$(a{e}_0+b{e}_1+c{e}_2)\wedge (x{e}_0+y{e}_1+z{e}_2)$$

in terms of the two-forms $e_i\wedge e_j$ give the cross product in ${\mathbb{R}}^3$.

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