# The Unapologetic Mathematician

## Inner Products of Vector Fields

Now that we can define the inner product of two vectors using a metric $g$, we want to generalize this to apply to vector fields.

This should be pretty straightforward: if $v$ and $w$ are vector fields on an open region $U$ it gives us vectors $v_p$ and $w_p$ at each $p\in U$. We can hit these pairs with $g_p$ to get $g_p(v_p,w_p)$, which is a real number. Since we get such a number at each point $p$, this gives us a function $g(v,w):U\to\mathbb{R}$.

That this $g$ is a bilinear function is clear. In fact we’ve already implied this fact when saying that $g$ is a tensor field. But in what sense is it an inner product? It’s symmetric, since each $g_p$ is, and positive definite as well. To be more explicit: $g_p(v_p,v_p)\geq0$ with equality if and only if $v_p$ is the zero vector in $\mathcal{T}_pM$. Thus the function $g(v,v)$ always takes on nonneative values, is zero exactly where $v$ is, and is the zero function if and only if $v$ is the zero vector field.

What about nondegeneracy? This is a little trickier. Given a nonzero vector field, we can find some point $p$ where $v_p$ is nonzero, and we know that there is some $w_p$ such that $g_p(v_p,w_p)\neq0$. In fact, we can find some region $U$ around $p$ where $v$ is everywhere nonzero, and for each point $q\in U$ we can find a $w_q$ such that $g_q(v_q,w_q)\neq0$. The question is: can we do this in such a way that $w_q$ is a smooth vector field?

The trick is to pick some coordinate map $x$ on $U$, shrinking the region if necessary. Then there must be some $i$ such that

$\displaystyle g_p\left(v_p,\frac{\partial}{\partial x^i}\bigg\vert_p\right)\neq0$

because otherwise $g_p$ would be degenerate. Now we get a smooth function near $p$:

$\displaystyle g\left(v,\frac{\partial}{\partial x^i}\right)$

which is nonzero at $p$, and so must be nonzero in some neighborhood of $p$. Letting $w$ be this coordinate vector field gives us a vector field that when paired with $v$ using $g$ gives a smooth function that is not identically zero. Thus $g$ is also nonzero, and is worthy of the title “inner product” on the module of vector fields $\mathfrak{X}(U)$ over the ring of smooth functions $\mathcal{O}(U)$.

Notice that we haven’t used the fact that the $g_p$ are positive-definite except in the proof that $g$ is, which means that if $g$ is merely pseudo-Riemannian then $g$ is still symmetric and nondegenerate, so it’s still sort of like an inner product, like an symmetric, nondegenerate, but indefinite form is still sort of like an inner product.