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Is the identifiaction between a vector space $V$ and its first dual $V^*$ functorial ? How can I see either positive or negative answer ?

EDIT At least with the use of a scalar product!

user122424
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    To be precise, you also need to specify how the morphisms are mapped. There is a natural way to define this mapping such that you get a (contravariant) functor. – ljfa Aug 10 '22 at 10:52
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    But this is not an "identification" of $V$ with $V^$. And when $V$ is infinite-dimensional, $V$ is not even isomorphic to $V^$. – GEdgar Aug 10 '22 at 10:57
  • @GEdgar Good. Stick with finite dimensionals $V$. – user122424 Aug 10 '22 at 11:06
  • @GEdgar But please see the EDIT. – user122424 Aug 10 '22 at 11:25
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    @user122424 what do you mean by the scalar product? Do you mean $v \cdot w = v^T w$. In that case the answer is yes because we can see from the definition that defining a scalar product is equivalent to defining a map $(\cdot)^T : V \to V^*$ given by $v \mapsto v^T$. – Charles Hudgins Aug 10 '22 at 11:33
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    In other words, we have the rather uninspiring answer that if we specify a functor from $V$ to $V^$ then the relationship between $V$ and $V^$ is functorial. But there is no natural such functor. The map $v \mapsto v^T$ is basis dependent. – Charles Hudgins Aug 10 '22 at 11:35
  • @CharlesHudgins I think that you've solved my entire problem. – user122424 Aug 10 '22 at 11:35
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    @user122424 it occurs to me that the specification of such a functor for general finite dimensional vector spaces probably requires the axiom of choice. Essentially we have to pick a basis for each vector space and define our functor as the usual row-column swap once we've written each vector down as components in the chosen basis for its vector space. Picking a basis for each vector space is highly noncanonical. – Charles Hudgins Aug 10 '22 at 11:38
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    As a final comment, note that there is a functor from vector spaces $V$ to their double duals $V^{}$ (note however that $V$ and $V^{}$ are only isomorphic in the finite dimensional case). Set $F(v)w = w(v)$ for $v \in V$ and $w \in V^*$. You still have to do a little checking and unwinding of definitions, but that's the idea. No finite dimensional restriction necessary. – Charles Hudgins Aug 10 '22 at 11:44
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    Sorry, final final comment. Any functor from $V$ to $V^*$ should be a contravariant functor, as you can check. – Charles Hudgins Aug 10 '22 at 12:01
  • Does this answer your question Relationship between a vector space V and its dual space V* – SoG Aug 11 '22 at 16:27

1 Answers1

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For a real inner product space $( V, \langle\cdot,\cdot\rangle)$ there is a natural map from $V$ to $V^*$. Namely, vector $u \in V$ goes to linear functional $\phi_u \in V^*$ defined by $\phi_u(x) = \langle x,u\rangle$. In case $V$ is finite-dimensional, this is an isomorphism from $V$ to $V^*$. Make it a (contravariant) functor as usual: if $T : V_1 \to V_2$ is a linear transformation, then $T^* : V_2 \to V_1$ satisfies $$ \langle T(x),u \rangle = \langle x,T^*(u)\rangle . $$ That is, with functionals as above, $T^* : V_2^* \to V_1^*$ is $$ T^*(\alpha) = \alpha\circ T,\quad\text{for }\alpha \in V_2^* $$

EDIT: Does this n.t. commutativity square make sense ? How can I justify that it does and that it really commutes ?

$$\begin{array}{ccc} V_1 & \rightarrow & V_1^* \\ {T} \downarrow && \ \ \ \uparrow {T^*}\\ V_2 & \rightarrow & V_2^* \end{array}\tag1$$

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No, $(1)$ is certainly wrong. If $V_1$ is $\mathbb R^2$ with the Euclidean norm, then we can think of $V_1$ as $1 \times 2$ matrices, and the correspondence $^*$ lets us think of $V_1^*$ as $1\times 2$ matrices. The horizontal arrows map each vector to its transpose. If $T : V_1 \to V_2$ is represented by a $2 \times 2$ matrix acting on the left, then $T^*$ is the transpose matrix acting on the right. For example, if $T = 2I$, twice the identity, then in $(1)$ the arrow across the top is the transpose, while the composition around the other three sides is $4$ times the transpose.

GEdgar
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  • Could you please state what does it mean now that ${}^$ is functorial* ? At best, if you could draw the fully labelled commutative square. – user122424 Aug 10 '22 at 14:04
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    I guess "functorial" would mean $1_{V}^* = 1_{V^}$ and $(S\circ T)^ = T^\circ S^$. – GEdgar Aug 10 '22 at 16:12
  • I think you are right. How would you prove this is the case for your ${}^*$ ? – user122424 Aug 10 '22 at 19:00
  • Why not try to do it yourself? – GEdgar Aug 10 '22 at 21:05
  • I can prove $(S\circ T)^=T^\circ S^$ but what I need is a commutative naturality square*. – user122424 Aug 11 '22 at 08:22
  • Please see my EDIT to your question. There I've summarized all my doubts about the problem. In particular instead of $V_1$ I should have written $F(x)$ and similarly $G(x),F(y)$ and $G(y)$ for some functors $F$ and $G$ but I do not know which I'm rather confused ... – user122424 Aug 11 '22 at 15:29
  • The OP is about "functorial", not about "naturality". – GEdgar Aug 11 '22 at 16:17
  • I thought that functorial means naturality square commutes for every functors images of any map $x\to y$ ?? – user122424 Aug 11 '22 at 16:49