A unitary space, interpreted as a Euclidean space

Question

Let $(V, \gamma)$ be a $n$-dimensional unitary space. Let $V_{\mathbb{R}}$ be the vector space $V$, interpreted as a $2n$-dimensional $\mathbb{R}$-vector space.

I first want to show that $(V_{\mathbb{R}}, \gamma_{\mathbb{R}})$ is a Euclidean space with the transformation:

$$\gamma_\mathbb{R}: V_\mathbb{R} \times V_\mathbb{R} \to \mathbb{R}, \gamma_\mathbb{R}(x, y) = Re(\gamma(x, y))$$

Also, I want to show that $||v||_\gamma = ||v||_{\gamma_\mathbb{R}}$, and that, in case $(v_1, ..., v_n)$ is an orthonormal basis of $V$, $(v_1, i v_1, ..., v_n, i v_n)$ is an orthonormal basis of $V_\mathbb{R}$.

Answer 1

The bilinearity of $\gamma_{\mathbb{R}}$ apparently follows from the sesquilinearity of $\gamma$,

Correct. Since $\gamma$ is sesquilinear, it is $\mathbb{R}$-bilinear, and composition with an $\mathbb{R}$-linear map preserves $\mathbb{R}$-bilinearity. $\operatorname{Re}\colon \mathbb{C}\to \mathbb{R}$ is $\mathbb{R}$-linear.

but I can't see why $\gamma_{\mathbb{R}}$ is positive definite

Because $\gamma$ is. For $x\neq 0$, we have $0 < \gamma(x,x) = \operatorname{Re} \gamma(x,x) = \gamma_{\mathbb{R}}(x,x)$.

That incidentally also immediately shows $\lVert v\rVert_\gamma = \sqrt{\gamma(v,v)} = \sqrt{\gamma_{\mathbb{R}}(v,v)} = \lVert v\rVert_{\gamma_{\mathbb{R}}}$ for all $v\in V$.

The elements of $V_{\mathbb{R}}$ are

precisely the elements of $V$. We change nothing with the elements, all we do is "forget" that we can multiply with complex scalars, and restrict our attention to multiplication with real scalars. The underlying set of $V_{\mathbb{R}}$ and the additive group structure are the same as those of $V$. Only scalra multiplication is restricted from $\mathbb{C}\times V \to V$ to $\mathbb{R}\times V \to V$.

If $(v_1,\dotsc, v_n)$ is an ONB of $V$, then we have

$$\lVert v_k\rVert_{\gamma_{\mathbb{R}}} = \lVert v_k\rVert_\gamma = \lVert iv_k\rVert_\gamma = \lVert iv_k\rVert_{\gamma_{\mathbb{R}}} = 1$$

for $1 \leqslant k \leqslant n$, so the norms are right. Now we need to see that the vectors are mutually orthogonal. For $k\neq j$, we have

$$\gamma_{\mathbb{R}}(i^m v_k, i^\ell v_j) = \operatorname{Re} \gamma(i^m v_k, i^\ell v_j) = \operatorname{Re} \bigl(i^{m-\ell}\gamma(v_k,v_j)\bigr) = \operatorname{Re} \bigl(i^{m-\ell}\cdot 0) = 0,$$

so the orthogonality for $v_k, v_j$, $v_k, iv_j$, $iv_k, v_j$ and $iv_k, iv_j$. What is missing is the orthogonality of $v_k, iv_k$. But

$$\gamma_{\mathbb{R}}(v_k,iv_k) = \operatorname{Re} \gamma(v_k,iv_k) = \operatorname{Re} (-i\gamma(v_k,v_k)) = \operatorname{Re} (-i) = 0.$$