# Problem with definition of complexification $V \otimes_{\mathbb{R}}\mathbb{C}$

Let $V$ be a finite dimensional real vector space. According to Wikipedia, the complexification of $V$ is defined to be $$V \otimes_{\mathbb{R}}\mathbb{C}$$ This can be made into a complex vector space by defining complex scalar multiplication by $$\lambda( v \otimes z) := v\otimes (\lambda z) \qquad v \in V,\lambda,z \in \mathbb{C}$$ Hence we define it on simple tensors only. So one has to check that it fulfills the axiom for scalar multiplication. Take another vector $u \otimes w$, where $u \in V$ and $z \in \mathbb{C}$. It is well-known, that $v \otimes z + u \otimes w$ must not be simple. So how can one show $$\lambda(v \otimes z + u \otimes w)$$ ?

I mean, $V \otimes_{\mathbb{R}}\mathbb{C}$ is a real vector space, and with the above we want to make it into a complex one. How exactly does one show the distributive property?

• $v \otimes \lambda z + u \otimes \lambda w$ – Randall Sep 2 '17 at 23:09
• If you take for granted that the scalar multiplication is well defined and that $V \otimes_{\mathbb{R}} \mathbb{C}$ is a complex vector space, then it suffices to say what scalar multiplication does to simple tensors, by the distributive property. – D_S Sep 2 '17 at 23:09
• So are you asking how to make a well defined scalar multiplication function $$\mathbb{C} \times V \otimes_{\mathbb{R}} \mathbb{C} \rightarrow V \otimes_{\mathbb{R}} \mathbb{C}$$ ? – D_S Sep 2 '17 at 23:10
• @D_S Yes, exactly, since $V \otimes_{\mathbb{R}}\mathbb{C}$ only is a real vector space. But if we define it like Wikipedia only on simple tensors, how do we get the distributive property? This is the thing I do not see right now. – TheGeekGreek Sep 2 '17 at 23:13
• You force the distributive property. – Randall Sep 2 '17 at 23:15

You can define scalar multiplication using the universal property of tensor products. Let $K = \mathbb{R}, L = \mathbb{C}$. To simplify notation, let me write $V \otimes W$ for $V \otimes_K W$.

If $V$ and $W$ are vector spaces over $K$, then as you know, $V \otimes W$ is also a vector space over $K$. We have a canonical map $V \times W \rightarrow V \otimes W$ defined by $(v,w) \mapsto v \otimes w$.

If $M$ is any vector space over $K$, and $f: V \times W \rightarrow M$ is any $K$-bilinear map, then there exists a unique $K$-linear transformation $\overline{f}: V \otimes W \rightarrow M$ such that $\overline{f}(v \otimes w) = f(v,w)$ for any $v \in V, w \in W$. This is the fundamental property of tensor products which you gives you a well defined scalar multiplication.

The point is that for each $\lambda \in L$, the map $V \times L \rightarrow V \otimes L$ given by $(v,c) \mapsto v \otimes c \lambda$ is $K$-bilinear. So there exists a unique $K$-linear transformation $T_{\lambda}: V \otimes L \rightarrow V \otimes L$ defined on generators by $T_{\lambda}(v \otimes c) = v \otimes c\lambda$.

Now you have a well defined scalar multiplication function

$$L \times (V \otimes L) \rightarrow V \otimes L$$

$$(\lambda, x) \mapsto T_{\lambda}(x)$$

and you can easily check that it satisfies the required properties. For example, to get the distributive property: for any $x, x' \in V \otimes L$ and any $\lambda \in L$

$$\lambda \cdot (x + x') = T_{\lambda}(x+x') = T_{\lambda}(x) + T_{\lambda}(x')= \lambda \cdot x + \lambda \cdot x'$$

and with this done, you can now just check all the other properties on simple tensors $v \otimes c$.

• That is exactly what I was looking for! Thanks. Honestly, I think I was confused by Wikipedias definition on simple tensors only without mentioning any linear extension or something like this. Then it is impossible to sho distributivity or things like that. – TheGeekGreek Sep 2 '17 at 23:49
• Since you are just dealing with vector spaces, there are some ways around what I did. Like if $v_i$ and $w_j$ are bases for $V$ and $W$, you can define $V \otimes W$ to be the vector space with formal basis $v_i \otimes w_j$ and then do everything ad hoc. But when you generalize this procedure to modules over rings, you really need to use the universal property to get anything done. – D_S Sep 2 '17 at 23:51
• This is a good answer. Things with tensor products got easier for me once I learned to think of them in terms of the universal property and not individualized elements. – Randall Sep 3 '17 at 0:15

There are two ways to define the complexification of a real vector space $V$: as a tensor product $\mathbf C \otimes_{\mathbf R} V$ (or $V \otimes_{\mathbf R} \mathbf C$, as you write it) with scaling by complex numbers on an elementary tensor $z \otimes v$ being applied to the complex tensorand $z$, or as a direct sum $V \oplus V$ that is thought of informally as $V \oplus iV$ with scaling defined as the symbols suggest: for $a, b \in \mathbf R$, $$(a + bi)(v+iw) = av + ibv + iaw - bw = (av - bw) + i(bv + aw).$$ A comparison of these two approaches, showing they really define the same thing, is in http://www.math.uconn.edu/~kconrad/blurbs/linmultialg/complexification.pdf.

By defining scalar multiplication, you are defining it on a basis for the complexification. Extend it linearly, then check that it is well-defined.

This, of course, is how it should work and given that it is often used as the definition, I would suppose that it has to work.

You have to consider that the tensors $v\otimes_\mathbb{R} w$ with $v\in V, w\in\mathbb{C}$ are the generators of $V\otimes_\mathbb{R}\mathbb{C}$, so that if you see someone defining some application which can be extended linearly, do it, with that spirit we have that: $$\lambda\cdot\sum_{\substack v\in V, w\in\mathbb{C} \\ finite\ sum}v\otimes_\mathbb{R}w:= \sum_{\substack v\in V, w\in\mathbb{C} \\ finite\ sum}v\otimes_\mathbb{R}\lambda w.$$