# Show that $g_{ij}v^i = v_j$

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.

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.
• 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. Mar 5, 2020 at 1:58