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.

September 30, 2011 - Posted by John Armstrong | Differential Geometry, Geometry

2 Comments »

[…] next step after using a metric to define an inner product on the module of vector spaces over the ring of smooth functions is to flip it around to the […]

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[…] Armstrong: (Pseudo)-Riemannian Metrics, Isometries, Inner Products on 1-Forms, The Hodge Star in Coordinates, The Hodge Star on Differential Forms, Inner Products on Differential […]

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