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I know for a vector space $V$ over the real or complex numbers, there exists a canonical embedding into its double dual $V^{**}$, and if $V$ is given an inner product, then there is a canonical embedding from $V$ into $V^*$.

However, I was not sure if there is any canonical embedding from $V^*$ into $V^{**}$ that is compatible with the above emebeddings, i.e. if $$\sigma:V \rightarrow V^{**}, \sigma(v)(f)=f(v)$$ $$\iota:V \rightarrow V^*, \iota(v)(w)=\langle w,v \rangle$$ are injective linear map.

So my question is: Is there a canonical (does not depend on basis) injective linear map $\mu: V^* \rightarrow V^{**}$ such that $$\sigma = \mu \circ \iota$$ I try to build an inner product on $V^*$ based on the inner product on $V$ but to no avail.

  • Your whole question is about definitions. What is the definition of independent of basis ? I think the answer is no so you need a way of disproving the statement. There is a categorical definition, but here is a second question, which maps $f:V\to W$ for inner product spaces do you consider ? You can only take the orthogonal to have the imbedding $i$ be canonical. Course if your on the category of finite spaces then the answer is yes, cause all maps are isomorphisms. – Rene Schipperus Dec 17 '17 at 04:26
  • About canonical map, I mean the categorical definition (the one with natural transformation), though for $\iota: V \rightarrow V^$ is not technically so. I probably not get what you mean in the later half, but I look for general linear map (not just specific kind of map between inner product space), and I had thought that $\mu$ must be induced from some inner product structure on $V^$ just like $\iota$ (I probably need to remove "canonical" from the maps that use inner product) –  Dec 17 '17 at 04:52
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    related $V$ is isomorphic to $V^{\ast\ast}$, the double dual space of $V$. – glS Aug 04 '20 at 07:35

1 Answers1

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When $V$ is finite dimensional, $\iota$ is an isomorphism, so you can just take $\mu=\sigma\circ \iota^{-1}$. In general, however, no such canonical map $\mu$ exists.

As evidence for this, consider the case that $V=\mathbb{R}^{\oplus\mathbb{N}}$ with the usual dot product as the inner product. Then $V^*$ can naturally be identified with $\mathbb{R}^{\mathbb{N}}$, and the image of $\iota$ is the subspace $\mathbb{R}^{\oplus \mathbb{N}}$ of sequences of finite support. Note now that it is consistent with ZF that the dual of $\mathbb{R}^\mathbb{N}$ is just $\mathbb{R}^{\oplus \mathbb{N}}$. In that case, there cannot exist any linear injection $V^*\to V^{**}$ at all, essentially since $V^{**}$ is countable-dimensional and $V^*$ is not (though it takes a bit of work to make this argument rigorous in the absence of choice). So, it is not possible to prove that such a map $\mu$ exists without using the axiom of choice.

Eric Wofsey
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  • Sorry for taking long enough to accept your answer because I'm getting stuck at why ${ s \in \mathbb{R}^B: \mu(s) \text{ vanishes on } \mathbb{R}^{\oplus B} }$ is a complement subspace to $\mathbb{R}^{\oplus B}$. Anyway, thank you for great answer. –  Aug 26 '18 at 23:05
  • Hmmm, revisiting this now, it seems that claim was wrong. I've replaced the argument with a different one (which has a somewhat weaker conclusion). – Eric Wofsey May 12 '19 at 05:02