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If $(V, \langle \cdot$ , $ \cdot\rangle)$ is an inner product space with dual $V^*$ then there is a natural dual pairing $\langle \cdot$ , $ \cdot \rangle ^*: V^* \times V \rightarrow \mathbb K$ given by $\langle f,v \rangle ^*= f(v)$.

Also any inner product space can be put in a dual pairing with itself via the inner product on it.

What I want to know is if $V$ is its own dual (that is $V^* = V$) then is this natural pairing the same as the inner product on $V$? That is for all $u, v \in V$ is $\langle u, v \rangle = \langle u, v \rangle ^*$

Thanks!

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    $\begingroup$ I think it depends on the isomorphism from $V$ to $V^*$ $\endgroup$ – Exodd Jun 9 '15 at 13:48
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    $\begingroup$ Could you expand on what you mean by the same? $\endgroup$ – muaddib Jun 9 '15 at 13:51
  • $\begingroup$ well if $u, v \in V$ then $\langle u,v \rangle$ gives the value of the inner product of $u$ and $v$. Now if i think of $u$ as a functional then is this value equal to $u(v)$? $\endgroup$ – R_D Jun 9 '15 at 13:59
  • $\begingroup$ hmm, Im having a hard time unwinding the question. Maybe I would ask it as, given the natural pairing $\langle f, v\rangle^*$ does there exist an inner product on $v$ s.t. every $f$ in the dual space can be represented using this inner product. Then maybe you end up with something like en.wikipedia.org/wiki/Riesz_representation_theorem $\endgroup$ – muaddib Jun 9 '15 at 14:12
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    $\begingroup$ Ok. I'm not sure yet how to make it any clearer. Thank you for your comment though. $\endgroup$ – R_D Jun 9 '15 at 14:33
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An inner product is an isomorphism $V \overset{\sim}\rightarrow V^*,\ v \mapsto v^*$. Or in bracket notation it's $v \mapsto \langle \cdot,v\rangle$.

For this question to make sense, you need to be able to compare the inner product and the dual pairing. To make that comparison you need such an isomorphism, which by definition is a map $v \mapsto v^*$ that makes $\langle u,v\rangle = \langle u,v^* \rangle$ hold.

It follows that this question makes sense if and only if the answer is trivially yes. However, this doesn't always happen, such as when there's no $v \mapsto v^*$.

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Let $V$ be a vector space. You have than nondegenerated pairing $<,>:V^*\times V\rightarrow\mathbb{R}.$ $$<f,v>=f(v)$$ Additionaly let $(,):V\times V\rightarrow\mathbb{R}$ be inner product in $V.$ And it gives you injective map $$\psi:V\rightarrow V^*;\psi(w)(v)=(w,v).$$ Hence for every $v,w\in V$ you get that $$<\psi(w),v>=\psi(w)(v)=(w,v).$$ You assume that $V$ is $V^*.$ Supose the word "is" means that $\psi$ is actually an isomorphism. Hence fixing $f\in V^*$ and $v\in V$ you get that $f=\psi(w)$ for some $w.$ And so $$<f,v>=f(v)=\psi(w)(v)=(w,v)=(\psi^{-1}(f),v).$$

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Take, for example, $\mathbb{R}^2=X$ with the standard scalar product. His dual is isomorphic to it, but there isn't an unique isomorphism.

For example, call $e_1,e_2$ the canonical base of $X$, and call $e_1^*,e_2^*$ the elements of $X^*$ such that $$e_i^*(e_i)=1\qquad e_i^*(e_{3-i})=0$$ If you take the isomorphism $e_i\to e_{3-i}^*$, then $$e_i^*(e_i)=1\ne 0=<e_{3-i},e_i>$$

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  • $\begingroup$ Isn't isomorphic different from equal? I asked for the case when $V = V^*$. So I thought the isomorphism is just the identity map. Did I miss something? $\endgroup$ – R_D Jun 9 '15 at 14:07
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    $\begingroup$ The problem here is what you call "identity map". You are seeing the elements of $V$ as functions $V\to \mathbb R$. That means you are associating the elements to functions, but the problem is that there are a lot of ways you can do such a thing. Maybe your question is "there exists an association that gives me the properties of the inner product?" $\endgroup$ – Exodd Jun 9 '15 at 14:45

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