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I'm reading Frankel's The Geometry of Physics. On pg. xxxiv, he defines $$ g_{ij} = \sum_k(\partial x^k/\partial u^i)(\partial x^k/\partial u_j)= \langle\partial\mathbf{p}/\partial u^i,\partial\mathbf{p}/\partial u^j\rangle = \langle\partial_i,\partial_j\rangle $$ and claims that $\langle\mathbf{v},\mathbf{w}\rangle = g_{ij}v^iw^j$. I didn't see how the product $\langle,\rangle$ is defined, so I assume that $\langle\mathbf{v},\mathbf{w}\rangle = v_jw^j$. Then, this should mean that $g_{ij}v^iw^j = v_jw^j$, or $g_{ij}v^i=v_j$, but if so, I'm having trouble proving it. One of my attempts went like this,

$$ \begin{split} g_{ij}v^i &= \langle\partial_i,\partial_j\rangle(v^i)\\ &= \partial_{ik}\partial_j^k(v^i), \end{split} $$

but I can't make any conceptual sense out of it.

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It depends on how you define the term "metric." One way to think about it is that a metric $g$ (say on $n$-dimensional Euclidean space $\mathbb{R}^n$) is an $n \times n$ matrix $g = (g_{ij})_{1 \leq i,j \leq n}$ of real numbers; then you may associate to $g$ a pseudo-inner product $\langle-,-\rangle$ which associates to any pair $\mathbf{v}=(v^1,\dots,v^n), \mathbf{w} = (w^1,\dots,w^n) \in \mathbb{R}^n$ of vectors, the scalar $$ \langle \mathbf{v},\mathbf{w} \rangle := \begin{pmatrix} v^1 & \cdots & v^n \end{pmatrix}\begin{pmatrix} g_{11} & \cdots & g_{1n} \\ \vdots & \ddots & \vdots \\ g_{n1} & \cdots & g_{nn} \end{pmatrix}\begin{pmatrix} w^1 \\ \vdots \\ w^n \end{pmatrix}, $$ or more explicitly, $$ \langle \mathbf{v},\mathbf{w} \rangle := g_{ij}v^iw^j $$ in summation notation. In this case, this equation is a definition, nothing to prove about it. (Conversely you can define a metric to be such a product $\langle-,-\rangle$, at which point you can obtain the corresponding matrix $g$ by $g_{ij} := \langle \mathbf{e}_i,\mathbf{e}_j \rangle$ where $\mathbf{e}_1,\dots,\mathbf{e}_n$ is a basis for $\mathbb{R}^n$; the two viewpoints are equivalent.) Then we define the operation of lowering indices, $$ v_j := g_{ij}v^i, $$ which transforms vectors written contravariantly to vectors written covariantly. This allows us to write the product $\langle -,- \rangle$ using the more compact notation $$ \langle \mathbf{v},\mathbf{w} \rangle = v_jw^j. $$ Again there's nothing to prove here – this follows immediately from the definition of lowering indices.

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  • $\begingroup$ Thanks for the answer, but I've talked with my professors and realized that my assumption that $\langle\mathbf{v}, \mathbf{w}\rangle = v_jw^j$ was incorrect, since it ommitted basis vectors, and was also incorrect notationally since both are vectors. So actually $\langle\mathbf{v}, \mathbf{w}\rangle = v^iw^j\langle\partial_i, \partial_j\rangle = g_{ij}v^iw^j$ now makes sense. $\endgroup$
    – David
    Mar 5, 2020 at 1:58

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