Are these two topologies same

Question

Consider the vector space $V=\mathbb{C}^n$. Equip $V$ with the Euclidean norm: $$||(x_1,\dots ,x_n)||=\sqrt{\sum |x_j|^{2}}$$ For any endomorphism $T:V\rightarrow V$, define the norm of $T$ as $$||T||=\sup\lbrace||Tx||: ||x||=1\rbrace$$ Consider the set $M_n(\mathbb{C})=$ Set of all $n\times n$ matrices with entries from $\mathbb{C}$. The topology on $\mathbb{C}$ induces the product topology on $M_n(\mathbb{C})=\mathbb{C}^{n^2}$. This makes $M_n(\mathbb{C})$ a topological ring. Then $GL_n(\mathbb{C})=M_n(\mathbb{C})^{\times}$ is a topological group.

The book I am reading, says that the topology of $GL_n(\mathbb{C})$ is given by the norm $||\text{ }||$.

I have the following questions

1. How exactly the norm $||\text{ }||$ is inducing a topology on $GL_n(\mathbb{C})$ ?

2. How is this topology same as the subspace topology inherited from $\mathbb{C}^{n^2}$ ?

Answer 1

In general, there is only one (Hausdorff) topology on a finite-dimensional ($\mathbb{R}$ or $\mathbb{C}$-)vector space making it a topological vector space (RUDIN - Functional Analysis, Theorem 1.21)

In the particular case of $M_n(\mathbb{C})$ with the two norms you described, we can show that the two norms are equivalent. Let's denote by $\Vert\cdot\Vert_2$ the norm on $\mathbb{C}^N$.

Let $T\in M_n(\mathbb{C})$ and $[t_{ij}]$ the associated matrix on the canonical basis $\left\{e_1,\ldots,e_n\right\}$. Then for each $j$, $$\Vert T\Vert^2\geq \Vert Te_j\Vert_2^2=\sum_{i=1}^n|t_{ij}|^2$$ so $\Vert T\Vert^2\geq\frac{1}{n}\sum_{j=1}^n\sum_{i=1}^n|t_{ij}|^2=\frac{1}{n}\Vert T\Vert_2^2$, that is, $\Vert T\Vert_2\leq \sqrt{n}\Vert T\Vert$.

On the other hand, there is $x=(x_1,\ldots,x_n)$ such that $\Vert x\Vert_2=1$ and $\Vert T\Vert=\Vert Tx\Vert_2$< so using Cauchy-Schwarz, $$\Vert T\Vert^2=\Vert Tx\Vert_2^2=\sum_{i=1}^n\left|\sum_{j=1}^nx_jt_{ij}\right|^2\leq\sum_{i=1}^n\Vert x\Vert_2^2\sum_{j=1}^n|t_{ij}|^2=\Vert T\Vert_2^2$$ so $\Vert T\Vert\leq \Vert T\Vert_2$.

This shows that both norms on $M_n(\mathbb{C})$ induce the same topology. More specifically, they induce metrics on $M_n(\mathbb{C})$ which in turn induce the same topology. Then $GL_n(\mathbb{C})$ has a subspace topology, given by either norm (metric): $d(A,B)=\Vert A-B\Vert$ or $d_2(A,B)=\Vert A-B\Vert_2$.