My question is very straightforward. Let $V$ be a vector space over $\mathbb{R}$. Is there any practical difference between the complexification $V_{\mathbb{C}}$ and implicitly treating $V$ as a complex vector space rather than a real one? In applications, it might be instructive to make such distinctions but if I'm treating $V$ as an abstract vector space over $\mathbb{R}$, isn't $V_{\mathbb{C}}$ equivalent to treat $V$ as a $\mathbb{C}$-vector space?

  • $\begingroup$ If $V$ is not even a vector space over $\Bbb{C}$, how can $V_{\Bbb{C}}$ even be equivalent to $V$ in any sense (heuristic or rigorous)? $\endgroup$
    – peek-a-boo
    Apr 29, 2021 at 20:01
  • $\begingroup$ What I mean is: what is there a difference between complexifying a vector space and simply begin to treat it like if it was defined over $\mathbb{C}$? For instance, if before $\alpha v$ was defined for $\alpha \in \mathbb{R}$, it is now defined for $\alpha \in \mathbb{C}$. In truth, the complexification is a tensor product with the complex plane, but can we abuse notation and treat it like a complex vector space? $\endgroup$
    – MathMath
    Apr 29, 2021 at 20:04
  • $\begingroup$ Maybe a more rigorous question would be: are they naturally isomorphic, so they can be treated as the same thing? $\endgroup$
    – MathMath
    Apr 29, 2021 at 20:06
  • $\begingroup$ Not sure I understand: Let's say $V$ is a real $2$-dimensional abstract vector space, i.e., a coordinateless plane with a distinguished origin. To "view $V$ as a complex vector space" you add structure, an automorphism $J$ with $J^{2} = -I$, though you still have a real $2$-dimensional space. To "complexify $V$" you tensor with $\mathbf{C}$ over the reals, obtaining a real $4$-dimensional space. So, what exactly is being prospectively compared? $\endgroup$ Apr 29, 2021 at 20:55

2 Answers 2


I think some confusion is arising from the fact that most "natural" real vector spaces are the real points of some other "natural" complex vector space. For example, if one considers the canonical $n$-dimensional real vector space $\mathbb{R}^n$ consisting of $n\times 1$ column vectors, then this obviously sits inside of $\mathbb{C}^n$.

Other examples include subsets of function spaces, since $\{ f:X\rightarrow \mathbb{R}\}\subseteq \{f:X\rightarrow \mathbb{C}\}$, e.g., the collection of real polynomials of degree less than or equal to $n$ is a subset of the corresponding collection of complex polynomials.

So, perhaps it is instructive to see an example where the naive extension to $\mathbb{C}$ does not work.

Let $V = (0,\infty)$ and define $x+_V y:= xy$ and, for $r\in \mathbb{R}$, $r\cdot_V x := x^r$. Then it's not too hard (though it is somewhat tedious to do from first principles) to verify that $V$ is real vector space with $0_V = 1$ and where the additive inverse of $x$ is $\frac{1}{x}$.

For example, $(rs)\cdot_V x = x^{rs} = (x^r)^s = s\cdot_V(r\cdot_V x)$, verifying that axiom.

(All the vector space axioms are trivially verified by simply noting that the bijection $e^x:\mathbb{R}\rightarrow (0,\infty)$ can be used to transport the usual structure on $\mathbb{R}$ to $V$).

For the naive extension to a $\mathbb{C}$ vector space, we should, for example, define $i\cdot x = x^i$. But $x^i$ has infinitely many complex values, even if $x$ is real. Of course, you can say "for each $x$, pick one value of $x^i$", but then this choice isn't continuous, and it will make it quite hard to verify that all the necessary axioms are satisfied.

On the other hand, $V_\mathbb{C}:=V\otimes_\mathbb{R} \mathbb{C}$ makes sense for any real vector space $V$, and gives a uniform way of complexifying all real vector spaces simultaneously. It even has the advantage that when you apply it to "natural" real vector spaces, it gives you (something isomorphic to) the "natural" corresponding complex vector space.

  • $\begingroup$ Jason, I accepted your answer yesterday but I was in a rush and I'd like to add a comment expressing my gratitute. Your answer was amazing! $\endgroup$
    – MathMath
    Apr 30, 2021 at 13:32
  • $\begingroup$ @MathMath: Glad you like it! $\endgroup$ Apr 30, 2021 at 13:33
  • $\begingroup$ Jason: If I am understood your example rightly, you give an example that $V_\mathbb{C}:=V\otimes_\mathbb{R} \mathbb{C}$ isn't same as $V_{\Bbb R}$. But $V_{\Bbb C}$ is a $k$ dimensional vector space over $\Bbb C$ and $V_{\Bbb R}$ is a $k$ dimensional vector space over $\Bbb R$ and any two $k$ dimensional vector spaces are isomorphic if I am not mistaken. Isn't? $\endgroup$
    – C.F.G
    May 4, 2021 at 0:46

We probably can't treat $V$ as a complex vector space. For instance, $R^n$ isn't closed under multiplication by complex numbers as you get products with nonzero imaginary components. On the other hand, the complexification $V \bigotimes C$ is guaranteed to be a vector space. Though the tensor product may be unintuitive, the resulting vector space is usually exactly what you'd expect. For instance, the complexification of $R^n$ is $C^n$.


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