$iV$ is open for $V$ open in real TVS

Question

Let $E$ be a complex vector space. It induces a real vector space $E_0$. Suppose that on $E_0$ we have a topology compatible with the vector structure, ie. a real TVS. How do we prove that $E$ is a complex TVS with same topology?

What I know is this: We have maps $$ \mathbb C \times E \to \mathbb R \times \mathbb R \times E \to E \times E \to E \\ (x + iy, e) \mapsto (x, y, e) \mapsto (xe, ye) \mapsto xe + iye $$

What's needed is $V$ open $\Rightarrow$ $iV$ open (ie. scalar multiplication by $i$ is continuous; note the inverse is $-i$).

Alternatively, if the claim were to be false, a counter-example would be much appreciated.

Answer 1

Your result is false : multiplication by $i$ may not be continuous.

Let $E_1$ and $E_2$ be two real topological vector spaces, with same underlying set but two different topologies. Let $E = E_1 \times E_2$ with the natural struture of real topological vector space and with complex structure : $$i : E_1 \times E_2 \longrightarrow E_1 \times E_2, \quad (x_1,x_2) \longmapsto (-x_2,x_1).$$ Then $E$ is not a topological complex vector space.

Answer 2

Suppose that $(\lambda_j)$ is a sequence in $\mathbb C$ and $\{e_k\}$ a net in $E$ such that $\lambda_j\to\lambda$ and $e_j\to e$. We want to show that $\lambda_je_k\to\lambda e$.

If $\lambda_j=a_j+ib_j$ with $a_j,b_j\in\mathbb R$ we have $a_j\to a$ and $b_j\to b$. Using that $E$ is a complex vector space, $$ \lambda_je_k=a_je_k+ib_je_k\to ae+ibe=(a+ib)e=\lambda e, $$ using that $E_0$ is a (real) TVS.