Isomorphism between $\mathbb{C}^{n}$ and the Complexification of $\mathbb{R}^{n}$

Question

What is an easy way (without tensors) to show, that the complexification of the Euclidean space $\mathbb {R} ^{n}$ gives the unitary space $\mathbb {C} ^{n}$. Or more detailed: I'd like to show, that $\mathbb{C}^{n}$ is isomorphic to the complexification $\left(\mathbb{R}^{n}\right)_{\mathbb{C}}$ of $\mathbb{R}^{n}$. I thought of it like this: $\mathbb{R}^{n}$ has the base $$ e_{j}=(0, \ldots, 0,1,0, \ldots, 0) \in \mathbb{R}^{n}, \quad j=1, \ldots, n . $$ Since the $1$ at position $j$ is also a complex number, $e_{j}$ can likewise be regarded as an element of $\mathbb{C}^{n}$, so there is an isomorphism. Is this already an acceptable explanation?

---

Definition of Complexification without tensors:

Let $V$ be a vector space over the field of real numbers $\mathbb{R}$. The complexification of $V$ is the direct sum $$ V_{\mathbb{C}}=V \oplus V=V \times V $$ On the new space, the addition is componentwise $$ (x, y)+\left(x^{\prime}, y^{\prime}\right)=\left(x+x^{\prime}, y+y^{\prime}\right) $$ and the scalar multiplication with $\alpha+\beta i \in \mathbb{C}$ by $$ (\alpha+\beta i)(x, y)=(\alpha x-\beta y, \beta x+\alpha y) $$ is defined. This makes $V_{\mathbb{C}}$ a vector space over the body of complex numbers $\mathbb{C}$.

In analogy to the notation of complex numbers, for the pair $(x, y) \in V_{\mathbb{C}}$ one also writes $x+iy$

Answer 1

If you are familiar with adjoint functors and free vector spaces, this explanation will be easy. If not, it will be incomprehensible.

The complexification functor $C : \mathbb{R}-Vec \to \mathbb{C}-Vec$ is the left adjoint of the forgetful functor $U_3 : \mathbb{C}-Vec \to \mathbb{R}-Vec$.

Note that the free functors $F_1 : Set \to \mathbb{R}-Vec$ and $F_2 : Set \to \mathbb{C}-Vec$ are the left adjoints to the forgetful functors $U_1 : \mathbb{R}-Vec \to Set$ and $U_2 : \mathbb{C}-Vec \to Set$ respectively.

Now note that $U_1 \circ U_3 = U_2$. Therefore, taking left adjoints, we have $C \circ F_1 \cong F_2$. In other words, the free complex vector space over a set $S$ is the complexification of the free real vector space over the set $S$.

Now $\mathbb{R}^n$ is the free vector space on $n$ elements. Therefore, its complexification is the free complex vector space on $n$ elements - that is, $\mathbb{R}^n$.

Answer 2

My rhetorical question to you is, what is your characterization (not "definition") of "complexification" of a real vector space, so that you know that your "construction/definition" gives the same thing as other versions?

A slightly fancy, but, I think, very illuminating, way to describe the complexification $CV$ of a real vector space $V$ is as an "adjoint functor" to the forgetful functor $\Phi$ that takes a complex vector space $W$ and "forgets" that it has complex scalar multiplication, but only "remembers" that it has real scalar multiplication. Then "complexification" $C$ is an adjoint functor in the sense that $$ \mathrm{Hom}_\mathbb C(CV,W)\;\approx\; \mathrm{Hom}_\mathbb R(V,\Phi W) $$ for every complex vector space $W$, and that the isomorphism is "natural" in a fairly reasonable sense, that real and complex vector space homs fit together with it.

As usual, this proves uniqueness up to unique isomorphism, if the thing exists at all. :)

The typical tensor product (or Hom) construction of complexification is proven to fulfill this characterization.

Presumably the tensor-product-less construction also can be proven to do so. :)

Still/and, "extension of scalars" (not just about real/complex, but far more generally) for free modules also can be proven to behave the way you indicate. For similarly very-general reasons. :)