# Isomorphic as inner product spaces

what do we mean by a complex space $H$ is isomorphic to $\mathbb{R}^n$ as an inner product space?

Does that mean that the isomorphism conserve the inner product?

Essentially yes, although you have to be a little bit careful about what is meant by isomorphism. Note that any complex vector space can be thought of as a real vector space with twice the dimension (e.g. $\mathbb{C}$ is a 2-dimensional real vector space with (real) basis $\{1,i\}$), so we say that a complex vector space $H$ (of dimension $n$) is isomorphic to $\mathbb{R}^{2n}$ if they are isomorphic as real vector spaces. If $\varphi:H\to \mathbb{R}^{2n}$ is such an isomorphism then it is said to conserve the (complex) inner product $\langle , \rangle_H$ on $H$ if $$\langle \langle \varphi(x),\varphi(y)\rangle_{\mathbb{R}^{2n}} = Re(\langle x,y\rangle_H) \qquad \text{for all x,y\in H}$$ (where $\langle,\rangle_{\mathbb{R}^{2n}}$ denotes the standard inner product on $\mathbb{R}^{2n}$ and $Re$ means taking the real part).
For example, let $H=\mathbb{C}^n$ (with its standard inner product). Then the isomorphism $\varphi:\mathbb{C}^n\to\mathbb{R}^{2n}$ given by $$\varphi(x_1+iy_1, x_2+iy_2,\dots,x_n+iy_n) = (x_1,y_1,x_2,y_2,\dots,x_n,y_n)$$ preserves the inner product in the above sense, as you can easily check by computing both sides of the above equation.