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I am trying to understand more about the Bidualspace (or double dual space). The whole idea is that $V$ and $V^{**}$ are canonically isomorphic to one another, which means that they are isomorphic without the choice of a basis, which means there exists an isomorphism between them which does not depend on choosing some basis (on either of the spaces). (suggestion by @DonAntonio)


Let $V$ be a finite dimensional Vectorspace with Basis $\beta = \lbrace v_1, \dots , v_n \rbrace $ (the infinite dimensional case is discussed at Canonical Isomorphism Between $\mathbf{V}$ and $(\mathbf{V}^*)^*$). I do understand that although we make a choice of Basis here, it will later on somehow be obsolete (which I don't see why)

Define: \begin{align} \begin{matrix} V & \overset{\Phi}{\longrightarrow}& V^* & \overset{\Phi^*}{\longrightarrow}&V^{**} \\ \sum_{i=1}^n \lambda_i v_i & \longmapsto & \sum_{i=1}^n \lambda_i v_i^* & \longmapsto & \sum_{i=1}^n \lambda_i v_i^{**} \end{matrix} \end{align} Discussion: I know and have already shown that $V$ and $V^*$ are isomorphic (not canonically isomorphic!) to one another, meaning that $\lbrace v_1^*, \dots , v_n^* \rbrace$ defines a Basis for $V^*$. Although I did not show it I am at peace with the statement that the mapping $\Phi^*$ introduces another isomorphism between $V^*$ and $V^{**}$. It is the canonical isomorphism between $V$ and $V^{**}$ that bothers me.

My tutors reasoning: In the following I will use $\checkmark$ to highlight whether or not I understand something.

  • $v_i^{**} \in V^{**}=\hom(V^{*},k)  \checkmark$, sure nothing to add here
  • $v_i^{**} (\sum_{j=1}^n \lambda_j v_j^{*})=\lambda_i \checkmark$, should be completely analogous to $v_i^*(v_j)=\lambda_i$
  • $\Phi^* \circ \Phi (v) =: \iota_v$ $$ \iota: \begin{cases} V & \longrightarrow V^{**} \\ v & \longmapsto \iota_v \end{cases}$$ where $\iota_v( \varphi)= \varphi(v)$ and $\varphi \in V^*$ is a linear functional. My tutor said that $\iota$ is 'suddenly' a canonical isomorphism, independent of the choice of the Basis $\beta$ because $\Phi$ and $\Phi^*$ are isomorphisms. This is the step where I understand nothing about.

What happened next (two more equations): I told my tutor that I don't see why $\iota$ is independent of a basis choice, because at $\Phi^* \circ \Phi= \iota$ we still make a choice for a basis at $\Phi$ namely $\beta$. My tutor said that this seems to be the case at first sight and made two more calculations which I in fact "understand": $$ \Phi^* \circ \Phi \left( \sum_{i=1}^n \lambda_i v_i \right) ( \varphi) = \Phi^* \left( \sum_{i=1}^n \lambda_i v_i^* \right) ( \varphi) = \left( \sum_{i=1}^n \lambda_i v_i^{**} \right)(\varphi) \\ = \sum_{i=1}^n \sum_{j=1}^n \lambda_i \mu_i \underbrace{v_i^{**}(v_j^*)}_{= \delta_{ij}} = \sum_{i=1}^n \lambda_i \mu_i $$ I understand this calculation, it's mainly an application of the above introduced definitions and the Kronecker-Delta. Apparently I am supposed to have an "aha" moment here, which unfortunately didn't occur yet. The next calculation he did was $$ \varphi (v) = \left(\sum_{j=1}^n \mu_j v_j^{*} \right) \left( \sum_{i=1}^n \lambda_i v_i \right)= \sum_{j=1}^n \sum_{i=1}^n \mu_j \lambda_i \underbrace{v_j^*( v_i)}_{\delta_{ij}} = \sum_{i=1}^n \lambda_i \mu_i $$ which evidently gives the same result as above. I would appreciate some wording on how those two equations (which calculations I understand) help me to see why there appears to be a canonical isomorphism between $V$ and $V^{**}$

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    $\begingroup$ "Isomorphic without the choice of a basis" could probably be a little inaccurate. I'd rather go with "there exists an isomorphism between them which doesn't not depend on choosing some basis (on either of the spaces)." $\endgroup$ – DonAntonio Apr 21 '14 at 16:29
  • $\begingroup$ Thank you @DonAntonio, I did add your suggestion $\endgroup$ – Spaced Apr 21 '14 at 16:47
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The first thing you mention is that, if you fix a basis for $V$, you get a non-canonical isomorphism $V\cong V^*$. Similarly, fixing a basis for $V^*$, you get a non-canonical isomorphism $V^*\cong (V^*)^*=V^{**}$. The "magic" is that, when you compose these two isomorphisms, you get an isomorphism $V\cong V^{**}$ which of course depends on your choice of bases, but which is exactly equal to the canonical isomorphism $$\iota: V\to V^{**};\quad \iota(v)=\iota_v,$$ regardless of what choices you made to define $\Phi$ and $\Phi^*$. The two extra equations you give show that $$\Phi^*\circ\Phi = \iota.$$ What may confuse you is that, in order to show these two maps are equal, we choose bases. But this is normal: we need these bases to define $\Phi$ and $\Phi^*$ to start with. But note that the two extra equations hold regardless of what bases were chosen.

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  • $\begingroup$ Thanks for your answer @MTurgeon, I did ponder about it for a while now and feel already better with it. Do the two discussed equations also highlight that they hold independent of the choice of the bases because the $v_i$ "vanish" during the calculations? Also is the resulting scalar in $V^{**}$? $\endgroup$ – Spaced Apr 22 '14 at 17:21
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The canonical map $\theta$ from a vector space to its double-dual has a particularly clean pointwise forumula:

$$\theta(v)(\omega) = \omega(v)$$

where $v$ is a vector and $\omega$ is a covector (an element of $V^*$).

Since $\omega$ ranges over all covectors, this gives a pointwise definition of $\theta(v)$. And since $v$ ranges over all vectors, this in turn gives a pointwise definition of $\theta$.

This notation is a little strange at first, but it's quite useful once you're used to it. $\theta(v)$ is an element of $V^{**}$, and so it makes sense to evaluate $\theta(v)$ at $\omega$ to get a scalar.

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