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In my Algebraic Topology class, we have just defined Topological Groups. We are then given a few examples, one of which is the orthogonal group $O(n)$.

For this to be a Topological Group we require that both the group multiplication and the action of taking the inverse of an element in the group are continuous maps.

Using the definition of continuous which we usually use in Topology, this means that we require that the pre-images of open sets are open, and this made me think about what the open sets in $O(n)$ actually are, and I realised I don't know. What is the topology on $O(n)$ and how is it defined?

EDIT: I do not need to be convinced that the map is continuous. This was shown in the lectures by asserting that the entries of the matrices appearing after these operations are polynomial. My question is for interest's sake.

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You should see $O(n)$ as a subset of $M_{n\times n}(\mathbb{R})$. And there is a natural distance in this space:$$d\bigl((a_{ij})_{1\leqslant i,j\leqslant n},(b_{ij})_{1\leqslant i,j\leqslant n}\bigr)=\sqrt{\sum_{i,j=1}^n|a_{ij}-b_{ij}|^2},$$which induces a topology. Then the topology of $O(n)$ is the subspace topology.

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