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Given a complex vector space $\mathcal{V}$, its complex conjugate $\overline{\mathcal{V}} = \{ \overline{v} : v \in \mathcal{V} \}$ consists of the "same" set of points (according to a number of references...). I'm struggling to reconcile that with the following example:


Define $ \mathcal{V} = \mathrm{span}_{\mathbb{C}}\{(1,i)\} = \{ (\alpha + \beta i, -\beta + \alpha i) : \alpha,\beta \in \mathbb{R} \}\,. $
Then, $(1,i) \in \mathcal{V}$ and so $\overline{(1,i)} \in \overline{\mathcal{V}}$, but $\overline{(1,i)} = (1,-i) \notin \mathcal{V}$. Thus, some vectors in $\overline{\mathcal{V}}$ are not in $\mathcal{V}$ (and vice versa).

Edited to add proposed solution (based on comments): Complex conjugation on $\mathcal{V}$ can be (re)defined as $$ \overline{(\alpha + \beta i, \, -\beta + \alpha i)} = (\alpha - \beta i, \,\beta + \alpha i) \in \mathcal{V} \,. $$


Am I missing something? If $\mathcal{V}$ is a complex vector space, then how do we reconcile it as having different vectors than its complex conjugate? On the other hand, if $\mathcal{V}$ isn't a complex vector space, then what is it?

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marked as duplicate by Alex M., Don Thousand, José Carlos Santos linear-algebra Oct 14 '18 at 14:01

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migrated from mathoverflow.net Apr 3 '15 at 2:37

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  • $\begingroup$ Indeed, this a good question to ask your instructor. $\endgroup$ – Włodzimierz Holsztyński Apr 3 '15 at 2:19
  • $\begingroup$ What sources are saying this? $\endgroup$ – rschwieb Apr 3 '15 at 2:45
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    $\begingroup$ A complex vector space doesn't have a natural notion of complex conjugation in general, it is an additional piece of information. $\endgroup$ – Michael Albanese Apr 3 '15 at 2:49
  • $\begingroup$ @rschwieb -- the first source I was led to (from Wikipedia is The Spinorial Chessboard by Budinich and Trautman. $\endgroup$ – Adam Apr 3 '15 at 4:01
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    $\begingroup$ @Adam: Every complex vector space admits a complex conjugation and therefore a real structure. The point is that you have to choose a complex conjugation, and different complex conjugations give different real structures. $\endgroup$ – Michael Albanese Apr 3 '15 at 5:16
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Complex conjugation is not a well-defined operation on a complex vector space. Having an operation that behaves like complex conjugation is precisely having a real structure.

If $V$ is a complex vector space, then $\overline{V}$ is another complex vector space which you can construct from $V$. Its underlying real vector space is the same as that of $V$, but its complex structure is different: scalar multiplication on $\overline{V}$ is the conjugate of scalar multiplication on $V$. Hence there is a natural map $V \to \overline{V}$ of real vector spaces, but it is explicitly not a map of complex vector spaces.

Your example is not well-defined either. $V$ is just some $1$-dimensional subspace of $\mathbb{C}^2$ and that's it. $\overline{V}$ is some $1$-dimensional subspace of the conjugate $\overline{\mathbb{C}^2}$. They live in different complex vector spaces, so it doesn't make sense to ask whether they agree or not. If you want to identify $\mathbb{C}^2$ with $\overline{\mathbb{C}^2}$ then you need to pick a real structure on $\mathbb{C}^2$.

All of this becomes much clearer once you introduce essentially any amount of extra structure; for example, you might work with complex representations of a group $G$ rather than just complex vector spaces. It's again the case that for every representation $V$ there is a conjugate representation $\overline{V}$ which again is defined by conjugating scalar multiplication. But $V$ and $\overline{V}$ are generally not isomorphic at all! In other words, generally complex representations of a group do not admit a real structure (compatible with the group action).

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  • $\begingroup$ Also I think the obvious proof that every complex vector space admits a real structure requires the axiom of choice. I don't actually know whether it's true without it. $\endgroup$ – Qiaochu Yuan Apr 3 '15 at 6:41

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