Evaluation under tensor products

Question

Could anyone explain to me the highlighted step in this

calculation? Can I just exaluate the tenor product component wise?

The first step is basically $$(ev_{V}\otimes id_{V})\circ(id_{V}(v)\otimes coev_{V}(v))=(ev_{V}\otimes id_{V})(v^ie_{i}\otimes(\epsilon^{j}\otimes e_{j}))$$ and then the second equality would look like

$$(ev_{V}\otimes id_{V})(v^ie_{i}\otimes(\epsilon^{j}\otimes e_{j}))=(ev_{V}((v^ie_{i}\otimes(\epsilon^{j}\otimes e_{j})))\otimes id_{V}((v^ie_{i}\otimes(\epsilon^{j}\otimes e_{j}))))$$. But I think I'm wrong there.

Answer 1

$\DeclareMathOperator{\ev}{ev} \DeclareMathOperator{\coev}{coev} \DeclareMathOperator{\id}{id} \renewcommand{\eps}{\epsilon}$Yes, you are indeed wrong. This is really confusing, so maybe it helps to write out where all the maps are going from and to. Note that I will omit the subscript $V$ below. As defined in the text, $$\begin{split}\ev&\colon V\otimes V^\vee \to k,\quad &e_i\otimes\eps^j\mapsto\eps^j(e_i)=\delta^j_i \\\coev&\colon k\to V^\vee\otimes V,&1\mapsto \eps^i\otimes e_i\end{split}$$ which hopefully makes it clear why they are "inverses". The only issue is that the orders in which $V$ and $V^\vee$ appear in the tensor product are reversed, so in order to really get inverses you want to tensor with another copy of $V$ in the appropriate side, namely $$\ev\otimes\id\colon V\otimes V^\vee\otimes V\to k\otimes V\cong V,\quad e_i\otimes\eps^j\otimes w\mapsto \delta^j_i\otimes w$$ and similarly for $\id\otimes\coev$: $$\id\otimes\coev\colon V\otimes k\cong V\to V\otimes V^\vee\otimes V,\quad w\otimes 1\mapsto w\otimes\eps^i\otimes e_i.$$ So now you really have two mappings with opposite domains and codomains which you can actually compose. Note that $V\otimes k$ and $k\otimes V$ are both naturally isomorphic to $V$, by identifying the tensor product with the scalar product on $V$. In other words $v=v\otimes 1$, and in general $cv=v\otimes c$ for $c\in k$.

The only thing is, well, from what is written above it is still not trivial that they are "inverses" (i.e. compose to the identity on $V$) since the order of the tensor products is different. So instead we just have to directly verify. Using Einstein summation convention, $v=v^ie_i$ and we write it out: $$ \begin{split}(\ev\otimes\id)\circ(\id\otimes\coev)(v)&=(\ev\otimes\id)\circ(\id\otimes\coev)(v\otimes1)\\&=(\ev\otimes\id)(v^ie_i\otimes\eps^j\otimes e_j)\\ &=\ev(v^ie_i\otimes\eps^j)\otimes\id(e_j) \\&=\eps^j(v^ie_i)\otimes e_j\\&=v^i\delta^j_i\otimes e_j\\&=v^ie_i=v=\id(v)\end{split} $$ as desired, so we are done.

Answer 2

Yes, you can calculate the tensor product "component-wise", i.e. if $f,g$ are maps, then $(f \otimes g)(x \otimes y) = f(x) \otimes g(y)$. In your case we get that \begin{align} (\text{ev}_V \otimes \text{id}_V)(v^ie_i \otimes \epsilon^j \otimes e_j) &= (\text{ev}_V \otimes \text{id}_V)\left((v^ie_i \otimes \epsilon^j) \otimes e_j\right)\\ &= \text{ev}_V(v^ie_i \otimes \epsilon^j) \otimes \text{id}_V(e_j)\\ &= \epsilon^j(v^ie_i) \otimes e_j\\ &= v^i\epsilon^j(e_i) \otimes e_j = v^i \delta_i^j \otimes e_j = v^ie_i = v. \end{align} I hope this is helpful!