Let $V = \oplus_{n \geq 0} V_n$, where $V_0, V_1, \ldots, V_n, \ldots$ are finite dimensional vector spaces. Why we have \begin{align} V^* = \prod_{n \geq 0} V_n^*, \end{align} but not \begin{align} V^* = \oplus_{n \geq 0} V_n^*? \end{align} Here $V^*$ is the dual vector space of $V$ and $V_i^*$ is the dual vector space of $V_i$. Thank you very much.
Dual vector space.
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linear-algebra
2 Answers
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For example: if $\pi_n:V \to V_n$ denotes the projection map, and if we select non-zero maps $f_n \in V_n^*$ for each $n$, then the map $F \in V^*$ given by $$ F(x) = \sum_{n \geq 0} f_n(\pi_n(x)) $$ is not an element of $\bigoplus_{n \geq 0} V_n^*$. Note that this map is well defined over $\bigoplus V_n$, but not over $\prod V_n$.
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Let $W$ be any vector space. By definition of the direct sum (i.e. coproduct in the category of vector spaces), for every sequence of linear maps $T_n:V_n\to W$ there is a unique linear map $T:V\to W$ such that $T\circ i_n=T_n$ where $i_n:V_n\to V$ is the inclusion map.
Now let $S_n:W\to V_n^*$ be a sequence of linear maps. Taking the dual maps of this sequence we have $S_n^*:(V_n^*)^*\to W^*$, and since $(V_n^*)^*\cong V_n$ canonically we have a sequence $S_n^*:V_n\to W^*$. By assumption, there is a unique linear map $T:V\to W^*$ such that $T\circ i_n=S_n^*$, and its dual $T^*:W\to V^*$ satisfies $$(T\circ i_n)^*=i_n^*\circ T^*=S_n$$ uniquely. Since this is true for any $W$ and any sequence $S_n$, it follows that $V^*$ is a product of the spaces $V_n^*$ with projection maps $i_n^*:V^*\to V_n^*$.
Basically the idea here is that $$ \hom(\oplus A_n,B)=\prod\hom(A_n,B) $$ since the product and coproduct are dual constructions, and the direct sum is a coproduct.