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Suppose that $(V,\langle \cdot\rangle_1)$ and $(V,\langle \cdot\rangle_2)$ are two real inner product vector spaces of the same dimension $n$, is it true that the bijective isometry between them are the orthogonal group $O(n)$?

An isometry here means the map that preserves the inner product.

It seems that this is not $O(n)$, but is it still able to calculate the dimension of it?

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  • $\begingroup$ you probably mean $(V, \|\cdot\|_2)$ $\endgroup$ – amakelov Apr 18 '17 at 22:33
  • $\begingroup$ No: given two functions $V \rightrightarrows W$, there is no composition defined, so it's not a group. $\endgroup$ – Chappers Apr 18 '17 at 22:39
  • $\begingroup$ I modified it a littbit $\endgroup$ – z.z Apr 18 '17 at 22:42
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    $\begingroup$ No. Consider $\langle,\rangle_2 = 4\langle,\rangle_1$. $\endgroup$ – user251257 Apr 18 '17 at 22:50
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The set of isometries between $(V,\left< \cdot, \cdot \right>_1)$ and $(V, \left< \cdot, \cdot \right>_2)$ is not even a group with respect to composition because it doesn't have the identity mapping unless $\left< \cdot, \cdot \right>_1 = \left< \cdot, \cdot \right>_2$. However, the set of isometries of $(V, \left< \cdot, \cdot \right>)$ does form a group which is isomorphic to $O(n)$ (assuming $V$ is a real $n$-dimensional vector space).

If you are looking for more structure, the set of isometries between $(V_1,\left< \cdot, \cdot \right>_1)$ and $(V_2, \left< \cdot, \cdot \right>_2)$ is an $O(V_1,\left< \cdot, \cdot \right>_1)$ and $O(V_2, \left< \cdot, \cdot \right>_2$) torsor. In particular, once a base point is fixed you can identify it with the orthogonal group. It has a natural structure of a manifold whose dimension is the same as the dimension of the orthogonal group $O(n)$.

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  • $\begingroup$ Do you have a reference on this? $\endgroup$ – z.z Apr 18 '17 at 23:46
  • $\begingroup$ @z.z: Nope, sorry. But you can work this out yourself. $\endgroup$ – levap Apr 18 '17 at 23:49

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