Continuous function of GL(n,C)

Question

I was reading a book on Lie groups where they define a function $f: GL(n,C)->GL(n,C)$, by

$f(B)=B^\dagger B$

where $B \in GL(n,C)$, and $GL(n,C)$ denotes the general linear group over C (complex field).

They mention something like "it's easy to see $f$ is continuous". It doesn't seem very obvious to me, is there a simple way to see this? I am not very fluent in topology so I apologize if this question is very trivial. Thanks in advance.

By definition the components of $f(B)$ are polynomials in the entries $b_{ij}$ of $B$ and their conjugates $\bar b_{ij}$. — Travis Willse, Dec 01 '17 at 00:58
I do not think algebraic topology is a good tag for this question — D_S, Dec 01 '17 at 01:02

score 2 · Accepted Answer · answered Dec 01 '17 at 00:57

1 . The topology on $G = \textrm{GL}_n(\mathbb{C})$ is the subspace topology coming from $\mathbb{C}^{n^2}$, the $n^2$-fold product of copies of $\mathbb{C}$.

2 . If $Z$ is a subspace of a topological space $Y$, then a function $f: X \rightarrow Z$ is continuous if and only if $f$ is continuous as a map from $X$ to $Y$. This follows directly from the definition of the subspace topology. Similarly, if $S$ is a subspace of $X$, and $f: X \rightarrow Y$ is a continuous function, then the restriction of $f$ to $S$ is a continuous function from $S$ to $Y$.

3 . If $Y$ is a topological space, and $n$ is an integer, let $Y^n$ be the $n$-fold product of copies of $Y$. Let $\pi_i: Y^n \rightarrow Y$ be the function $(y_1, ... , y_n) \mapsto y_i$. A function $f: X \rightarrow Y^n$ is continuous if and only if each $\pi_i \circ f: X \rightarrow Y$ is continuous. This follows directly from the definition of the product topology.

Using these principles, you can reduce the problem to verifying that certain functions $\mathbb{C}^{n^2} \rightarrow \mathbb{C}$ are continuous.

Thank you. I'll look more into those principles and try to do the proof. — Cami77, Dec 01 '17 at 01:21

Continuous function of GL(n,C)

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