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If I am correct dot product is an example of inner product on coordinate space.

I wonder if for an arbitrary inner product space with base field being $\mathbb{R}$ or $\mathbb{C}$, there always exists a coordinate system so that the inner product becomes the dot product in coordinate? What is the name of the topic regarding this question?

Thanks and regards!

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    So, the more general topic is that of pull backs of 2-forms between manifolds.2011-08-02

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For finite dimensional spaces, the answer is "yes"; this is a consequence of the Gram-Schmidt orthonormalization process: every finite dimensional inner product space over $\mathbb{R}$ or over $\mathbb{C}$ has an orthonormal basis.

Now let $\mathbf{V}$ be an inner product space, and let $\mathbf{v}_1,\ldots,\mathbf{v}_n$ be an orthonormal basis. Then $T\colon\mathbf{V}\to \mathbf{F}^n$ given by $T\mathbf{v}_i = \mathbf{e}_i$ (i.e., $T$ maps each vector in $\mathbf{V}$ to its coordinate vector relative to the orthonormal basis $\mathbf{v}_1,\ldots,\mathbf{v}_n$) is an invertible linear transformation such that for all $\mathbf{x},\mathbf{y}\in\mathbf{V}$, $\langle \mathbf{x},\mathbf{y}\rangle = T\mathbf{x}\cdot T\mathbf{y}$, where the right hand side is the standard dot product on $\mathbf{F}^n$ ($\mathbf{F}=\mathbb{R}$ or $\mathbb{C}$).

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    In fancy terms, every finite dimensional inner product space is isometric to $\mathbf F^n$ (but not in a canonical way).2011-08-02