Let $V$ be a finite dimensional complex vector space.

I recently asked how to find a

Natural isomorphism between $\mathbb{C}\otimes_{\mathbb{R}}V$ and $V\oplus V.$

and got some very nice answers. In particular, I was told that the map

$$c\otimes v \mapsto (\Re(c)v,\Im(c)v)$$

is a complex linear isomorphism.

Here is the problem I am having with this:

As I understand it, the scalar multiplication defined on the extension of scalars is given by

$$c'(c\otimes v) = (c'c)\otimes v$$

and with this multiplication the above map is not complex linear. However, if I use $$c'(c\otimes v)=c\otimes (c'v)$$ as my scalar multiplication then the given map is indeed complex linear.

So how do I reconcile this with the definition of scalar multiplication in an extension of scalars?

Many thanks!

  • 1
    $\begingroup$ The isomorphism should actually be stated as being between $C \otimes V$ and $V \oplus \overline{V}$, where $\overline{V}$ is $V$ with the conjugate complex structure. This is isomorphic to $V$ but not naturally so, and this statement is the one that continues to make sense e.g. for vector bundles. $\endgroup$ – Qiaochu Yuan Aug 25 '16 at 18:29

In order to make things less confusing, let me set up some notation:

  1. Given a complex vector space $V$, I will write $\operatorname{For}(V)$ for the underlying real vector space ($\operatorname{For}$ stands for "forgetful" and this operation is called in general restriction of scalars).
  2. When convenient, I will think of a complex vector space as a pair $(W,J)$ where $W$ is a real vector space and $J \colon W \rightarrow W$ is a real linear map satisfying $J^2 = -\operatorname{Id}_W$. Such a map is called a linear complex structure on $W$ and it defines the structure of a complex vector space on $W$ by the rule $(a + ib)w := aw + bJw$. Going the other direction, a complex vector space $V$ gives a pair $(\operatorname{For}(V), J_V)$ where $J_V \colon \operatorname{For}(V) \rightarrow \operatorname{For}(V)$ is defined by $J_V(v) := iv$.
  3. The complexification (extension of scalars) of a real vector space $W$ is defined to be $W_{\mathbb{C}} := (\operatorname{For}(\mathbb{C}) \otimes_{\mathbb{R}} W, i \otimes \operatorname{Id}_W)$ where $i = J_{\mathbb{C}} \colon \operatorname{For}(\mathbb{C}) \rightarrow \operatorname{For}(\mathbb{C})$ is the natural complex structure on $\mathbb{C}$. That is, the complex multiplication on $W_{\mathbb{C}}$ is defined using the complex structure on $\mathbb{C}$ acting "from the left" (this is the only sensible choice as $W$ is only a real vector space).

Now, let us assume that $V$ is a complex vector space. We can form the real vector space $\operatorname{For}(\mathbb{C}) \otimes_{\mathbb{R}} \operatorname{For}(V)$ and then endow it with two possible complex structures:

  1. One is $i \otimes \operatorname{Id}_V$ which comes from the complex structure on $\mathbb{C}$. The resulting complex vector space $(\operatorname{For}(\mathbb{C}) \otimes_{\mathbb{R}} \operatorname{For}(V), i \otimes \operatorname{Id}_V)$ is the one usually called the complexification of a complex vector space $V$ and using our previous notation, can be denoted by $\operatorname{For}(V)_{\mathbb{C}}$ (we forget about the complex vector structure of $V$, treat it as a real vector space and complexify it as before).
  2. The other is $\operatorname{Id}_{\mathbb{C}} \otimes J_V$ which comes from the complex structure on $V$. The resulting complex vector space $(\operatorname{For}(\mathbb{C}) \otimes_{\mathbb{R}} \operatorname{For}(V), \operatorname{Id}_{\mathbb{C}} \otimes J_V)$ is not usually called the complexification of $V$.

Using the terminology above, there is no natural complex isomorphism between $\operatorname{For}(V)_{\mathbb{C}}$ and $V \oplus_{\mathbb{C}} V$ which shouldn't be surprising as the construction of $\operatorname{For}(V)_{\mathbb{C}}$ doesn't use the complex structure on $V$ at all while the construction of $V \oplus_{\mathbb{C}} V$ certainly does.

On the other hand, $(\operatorname{For}(\mathbb{C}) \otimes_{\mathbb{R}} \operatorname{For}(V), \operatorname{Id}_{\mathbb{C}} \otimes J_V)$ is indeed isomorphic to $V \oplus_{\mathbb{C}} V$ (written in pair notation as $(\operatorname{For}(V) \oplus_{\mathbb{R}} \operatorname{For}(V), J_V \oplus J_V)$).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.