# Understanding extension of scalars

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!

• 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. – 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 $(\operatorname{For}(V) \oplus_{\mathbb{R}} \operatorname{For}(V), J_V \oplus J_V)$).