Introduction to tensor (4) how to define trace?

In this blog, we would like to define the trace $tr(S)$$tr(S)$ of $S\in {\mathrm{End}}_{R-\mathrm{Mod}}(R^n)$$S\in\mathrm{End}_{R-\mathrm{Mod}}(R^n)$​ from the tensor point of view.

Consider a natural transformation $\eta :{\mathrm{Hom}}_{R-\mathrm{Mod}}(-,P)\otimes N⟶{\mathrm{Hom}}_{R-\mathrm{Mod}}(-,P\otimes N)$$\eta:\mathrm{Hom}_{R-\mathrm{Mod}}(-,P)\otimes N\longrightarrow\mathrm{Hom}_{R-\mathrm{Mod}}(-,P\otimes N)$​

Denote them as $F,G$$F,G$.

$$\begin{array}{ccc}{\mathrm{Hom}}_{R-\mathrm{Mod}}(M,P)\otimes N& \stackrel{{\eta }_M}{\to }& {\mathrm{Hom}}_{R-\mathrm{Mod}}(M,P\otimes N)\\ F(f)\downarrow & & \downarrow G(f)& \\ {\mathrm{Hom}}_{R-\mathrm{Mod}}(M^{\prime },P)\otimes N& \underset{{\eta }_M^{\prime }}{\overset{}{\to }}& {\mathrm{Hom}}_{R-\mathrm{Mod}}(M^{\prime },P\otimes N)\end{array}$$

Here ${\eta }_M$$\eta_M$ (similarly, ${\eta }_{M^{\prime }}$$\eta_{M'}$) is defined as:

$$u\otimes n⟼(m\mapsto u(m)\otimes n)$$

It is easy to verify that this forms a natural transformation.

A good news is, when we restrict the source of $F,G$$F,G$ to category of finite rank free $R-$$R-$module, then $\eta$$\eta$ will become a natural isomorphism! Since we are considering category of finite rank free $R-$$R-$module, we only need to prove it for

$$M\cong R^n$$

But

$${\mathrm{Hom}}_{R-\mathrm{Mod}}(R^n,P)\otimes N\cong \prod _{i=1}^n{\mathrm{Hom}}_{R-\mathrm{Mod}}(R,P)\otimes N$$

and

$${\mathrm{Hom}}_{R-\mathrm{Mod}}(R^n,P\otimes N)\cong \prod _{i=1}^n{\mathrm{Hom}}_{R-\mathrm{Mod}}(R,P\otimes N)$$

Hence we only need to prove that

$${\mathrm{Hom}}_{R-\mathrm{Mod}}(R,P)\otimes N\cong {\mathrm{Hom}}_{R-\mathrm{Mod}}(R,P\otimes N)$$

But as we know the identity functor is representable.

$${\mathrm{Hom}}_{R-\mathrm{Mod}}(R,H)\cong H$$

Hence we only need to prove that

$$P\otimes N\cong P\otimes N$$

It is automatically true. We down! $◻$$\square$

Then let $P=R$$P=R$ we will get $M^{\vee }\otimes N\cong {\mathrm{Hom}}_{R-\mathrm{Mod}}(M,N)$$M^{\vee}\otimes N\cong\mathrm{Hom}_{R-\mathrm{Mod}}(M,N)$. Here $M^{\vee }$$M^{\vee}$ means the dual module of $M$$M$​.

$$m^{\ast }\otimes n⟼(v\mapsto m(v)n)$$

Now let us consider the tensor-hom adjoint again.

$${\mathrm{Hom}}_{R-\mathrm{Mod}}(M\otimes N,Q)\cong {\mathrm{Hom}}_{R-\mathrm{Mod}}(M,{\mathrm{Hom}}_{R-\mathrm{Mod}}(N,Q))$$

When $N,Q$$N,Q$ are free module, we have ${\mathrm{Hom}}_{R-\mathrm{Mod}}(N,Q)\cong N^{\vee }\otimes Q$$\mathrm{Hom}_{R-\mathrm{Mod}}(N,Q)\cong N^{\vee}\otimes Q$.

Hence we have

$${\mathrm{Hom}}_{R-\mathrm{Mod}}(M\otimes N,Q)\cong {\mathrm{Hom}}_{R-\mathrm{Mod}}(M,N^{\vee }\otimes Q)$$

If we consider the category of free $R-$$R-$module, or even category of $\mathbb{F}-$$\mathbb F-$ vector space, we get that

$-\otimes N$$-\otimes N$ is the left adjoint of $N^{\vee }\otimes -$$N^\vee\otimes -$.

Remark. Relation wuth matrix representation of linear map.

$$\begin{array}{ccccc}& & {\mathrm{Hom}}_{R-\mathrm{Mod}}(M,N)& & \\ & & \downarrow & & & \\ M^{\vee }\otimes N& \stackrel{}{\to }& M_{h,k}(R)\end{array}$$

Let $m^1,...,m^h$$m^1,...,m^h$ be the basis of $M^{\vee }$$M^{\vee}$ and $n_1,...,n_k$$n_1,...,n_k$ be the basis of $N$$N$​, $n^1,...,n^k$$n^1,...,n^k$ be the dual basis.

$$n^i(T{m}_j)n_i=n^i\circ T(m_j)n_i=T^{\prime }(n^i)(m_j)n_i=m(m_j)n_i$$

Here $T^{\prime }$$T'$ is the dual map of $T$$T$​​.

$$M^{\vee }\otimes N\cong N\otimes M^{\vee },\sum _{i,j}{a}_{i,j}{n}_i\otimes m^j⟼\sum _{i,j}{a}_{i,j}{E}_{i,j}$$

In particular, if $N=M$$N=M$, we get that $f:{\mathrm{End}}_{R-\mathrm{Mod}}(M)\cong M^{\vee }\otimes M$$f:\mathrm{End}_{R-\mathrm{Mod}}(M)\cong M^{\vee}\otimes M$​​​.

Let us define the pre-trace map $\tau$$\tau$ as follows.

$$\tau :M^{\vee }\otimes M\to R,v^{\ast }\otimes v⟼v^{\ast }(v)$$

When you have a basis of $M$$M$ such as $v_1,...,v_n$$v_1,...,v_n$ and its dual basis $v^1,...,v^n$$v^1,...,v^n$ then every element in $M^{\vee }\otimes M$$M^\vee\otimes M$ could be written as

$$\sum _{i,j}{a}_{i,j}{v}^i\otimes v_j$$

by the bilinear property.

The isomorphism between $M^{\vee }\otimes M\cong M_{n,n}(R)$$M^\vee\otimes M\cong M_{n,n}(R)$ is given by

$$v^i\otimes v_j\mapsto E_{i,j}$$

As you can see,

$$\tau :\sum _{i,j}{a}_{i,j}{v}^i\otimes v_j⟼\sum _i{a}_{i,i}$$

and the trace $Tr$$Tr$ is defined as

$$Tr:{\mathrm{End}}_{R-\mathrm{Mod}}(M)\stackrel{f}{\to }{M}^{\vee }\otimes M\stackrel{\tau }{\to }R$$