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For example, let $V$ be a complex vector space (vector space over C) of dimension 2 and $W=V\oplus V.$

Question:

1) Let $V^R$ denotes the decomplexification of V. Is W the complexification of $V^R$? I don't think so, because the complexification of $V^R$ should be given by the direct sum of real vector spaces. So, in this case, the complexification of $V^R$ should be $(V^R)^C = V^R \oplus V^R $.

Edit: I forgot to include that I am supposing $(V^R)^C$ has a linear complex structure that allows complex scalar multiplication. The complex structure is defined by the operator $J$ that is defined by $J(v,w)=(-w,v)$ for any $(v,w)$ element of $(V^R)^C$. So, for $(a+ib)$ in C, $(a+ib)(v,w)=(a+bJ)(v,w) = (av-bw, aw+bv)$.

2) If $W$ is not the complexification of $V^R$...

$W$ is clearly a complex vector space of dimension $4$. $(V^R)^C$ is a complex vector space of dimension 4 as well (since it has the same dimension as $V^R$). Are $W$ and $(V^R)^C$ isomorphic? if not, what is the relation between them?

I would appreciate any guidance. Thank you.

Cami77
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1 Answers1

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I think there is a misconception going on here. Taking the direct sum of a real vector space with itself $V \oplus V$ is still a vector space over $\mathbb R$. It is true that it has twice as many basis elements and can be thought of as $\mathbb R^{2n}$, but there is still no way to multiply complex numbers here, since scalar multiplication is not by complex numbers.

What you are looking for is $V \otimes \mathbb C$, which is a tensor product, and as a real vector space, it certainly has dimension $2n$ with basis "elements" $\{e_1, \dots e_n,ie_1, \dots ie_n\}$, but this is a misleading way of writing $e_i \otimes 1$ and $e_i \otimes i$, and the previous terminology is an abuse of notation.

From this point of view, we obtain a complex vector space by scalar multiplication defined by $c \times (v \otimes (a+bi)=v \otimes c(a +bi)$ with $c \in \mathbb C$.

In general, there is an identification $V \oplus i V \cong V \otimes \mathbb C=V^{\mathbb C}$,

but to answer your question: no, just taking a direct sum is not by itself a complexification.

Your last paragraph is also incorrect: $\mathrm \dim_{ \mathbb C}V^{\mathbb C}=\mathrm{dim}_{\mathbb R} V$, which is because every element can be written as a linear combination of complex numbers multiplied by vectors in the usual basis.

Andres Mejia
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  • Thank you. Yes I realize I forgot to include that $V^C$ needs to have a linear complex structure that allows complex scalar multiplication. The complex structure is defined by the operator $J$ that is defined by $J(v,w)=(-w,v)$ where $(v,w)$ is an element of $V^C$. So, for $(a+ib)$ in C, $(a+ib)(v,w)=(a+bJ)(v,w) = (av-bw, aw+bv)$. If $W$ would have such a complex structure as well, but still is a direct sum of a vector space over C, is $W$ related to $V^C$? – Cami77 Dec 08 '17 at 01:42
  • @Cami77 see my edit – Andres Mejia Dec 08 '17 at 01:51
  • Yes I made a mistake in the dimension of $V^C$. My main issue was the 'complexification' of a complex vector space, the idea was just confusing to me. This helped, thanks! – Cami77 Dec 08 '17 at 02:01
  • No problem. I cant resist the temptation to suggest you look up "extension of scalars" which generalizes this construction (at least if you are semi-acquainted with tensor products) – Andres Mejia Dec 08 '17 at 02:21
  • Thanks for the suggestion, I'll definitely do it. – Cami77 Dec 08 '17 at 05:50
  • I think there is a slightly misleading thing in this answer: The tensor product used for the complexification at first is over $\mathbb{R}$, but the one in the end you claim does not change $V$ is over $\mathbb{C}$ (otherwise it would indeed double the dimension). – Tobias Kildetoft Dec 08 '17 at 07:03
  • @AndresMejia I think Tobias is right. I found this link, Example 2.5 states that for the case of $C^R$ (complex numbers as a real vector space), $(C^R)^C$ is not isomorphic to $C$, since $dim_C (C^R)^C=dim_R (C^R)=2$ and $dim_C C=1$. http://www.math.uconn.edu/~kconrad/blurbs/linmultialg/complexification.pdf I am not very confident in tensor products, but the link might help you correct the last part of your answer. – Cami77 Dec 08 '17 at 08:12
  • Oh wow, that was certainly an oversight on my part. I’m not at my computer, so for now I’ve removed the last bit, but as Tobias mentioned, and it indeed does double the dimension. – Andres Mejia Dec 08 '17 at 14:35
  • That's fine. I was thinking (from my question) although $W$ is not equal to $(V^R)^C$ (V a complex vector space) in the sense that $W$ does't have/need a complex structure, $W$ is isomorphic to $(V^R)^C$ since $(V^R)^C \cong V^R\oplus iV^R \cong V \oplus V=W$. Right? – Cami77 Dec 08 '17 at 15:10
  • Andres Mejia and @Cami77 About 'no, just taking a direct sum is not by itself a complexification.', what about internal complexifications? – BCLC Jan 30 '20 at 08:18