Let $(M,g)$ be a Riemannian manifold and let $f \in C^{\infty}(M)$. Let $X$ be a smooth vector field on $M$. In smooth local coordinates $(x^i)$ on $M$, we can write $g = g_{ij} dx^i \otimes dx^j$ as well as $X = X^i \partial_{x^i}$. Now I have computed that $\langle\operatorname{grad} f, X \rangle_g = X^i\frac{\partial f}{\partial x^i}$ as follows. If we let $g^{ij}$ be the inverse matrix of $g_{ij}$ We have $$\begin{eqnarray*} \langle\operatorname{grad} f, X \rangle_g &=& \left\langle g^{ij} \frac{\partial f}{\partial x^i}\partial_{x^j} , X^k\partial_{x^k} \right\rangle_g\\ &=& g^{ij} \frac{\partial f}{\partial x^i}X^k\left\langle \partial_{x^j},\partial_{x^k} \right\rangle_g\\ &=& g_{jk}g^{ij} \frac{\partial f}{\partial x^i}X^k \\ &=& g_{kj}g^{ji}\frac{\partial f}{\partial x^i}X^k \hspace{1cm} (\text{by using that $g_{ij}$ is a symmetric matrix})\\ &=& X^i\frac{\partial f}{\partial x^i}. \end{eqnarray*}$$

My question is: Is there any geometric reason as to why $\langle \operatorname{grad} f, X \rangle_g$ should be independent of the metric $g$?


Absolutely. The exterior differential of a function, $df$, is a one-form which is defined independently of metric: $df(X) = X(f)$. In coordinates, $df = \partial_if dx^i$.

Choice of a nondegenerate metric induces an isomorphism between vector fields and $1$-forms by pairing: $g(X)(V) = \langle X,V\rangle$. In coordinates, this pairing corresponds to multiplying by the inverse of the metric, or "raising indices." This isomorphism is also known as a musical isomorphism: if $X$ is a vector field, then the one-form $X^\sharp$ is defined by $X^\sharp(V) = \langle X,V\rangle$.

The gradient $\operatorname{grad}(f)$ is defined to be the metric dual of the one-form $df$, i.e., $\operatorname{grad}(f) = (df)^\sharp$, so if you pair $\operatorname{grad}(f)$ with an arbitrary vector field $X$, you get $df(X) = X(f)$, which does not depend at all on the metric. The gradient is constructed so that, in pairing, all mentions of the metric cancel.

  • $\begingroup$ Thanks Neal for your answer. In our class we defined $\operatorname{grad}(f) := (df)^{\sharp}$ where $\sharp$ is the sharp operator. I expanded everything out because it was not clear to me that the inner product should not depend on $g$. $\endgroup$ – user38268 May 25 '13 at 14:43
  • $\begingroup$ Wait a sec @BenjaLim, the inner product does depend on $g$ simply because $g$ is the association of one inner product to each tangent space of $M$ in a smooth manner, so the inner product is given by $g$. $\endgroup$ – user1620696 May 25 '13 at 14:47
  • $\begingroup$ @BenjaLim The $\#$ operator is the isomorphism induced by pairing. I will edit this into my answer. $\endgroup$ – Neal May 25 '13 at 14:49
  • $\begingroup$ @BenjaLim Thanks. Basically, in your computation, you unwound the definition $\operatorname{grad}(f) = (df)^\sharp$. $\endgroup$ – Neal May 25 '13 at 14:54

Yes, the reason is simple. By definition, $\operatorname{grad}f$ is the vector field (metric-dependent) so that $\langle\operatorname{grad}f,X\rangle_g = df(X)$, and the right-hand side is independent of the metric. But the vector field itself does depend on the metric. :)

  • $\begingroup$ Oh one more thing: The only way I can see how to pass from the second last line to the last line is by writing out two matrices side by side. Is there a quick way to do this without writing out matrices? It will save me time in future. $\endgroup$ – user38268 May 25 '13 at 15:22
  • 2
    $\begingroup$ Well, that's what's going on, because $g^{ij}$ is defined to be the inverse matrix. So you have, of course, $g^{ik}g_{kj} = \delta^i_j$. In general, if $A$ and $B$ are matrices (i.e., $(1,1)$-tensors), $(AB)^i_j = A^i_k B^k_j$, whatever your row- and column- conventions are. $\endgroup$ – Ted Shifrin May 25 '13 at 15:26

Any Riemannian manifold can be alternatively described by gauge fields on a flat Euclidean manifold (it may take several gauge fields on overlapping subsets to completely describe the manifold, but it can be done).

Let the basis vectors of the flat Euclidean space be $e_1, e_2, \ldots$ with corresponding basis covectors $e^1, e^2, \ldots$. There is the usual partial derivative operator $\partial = e^1 \partial_1 + e^2 \partial_2 + \ldots$.

There is a gauge field $\underline h$, which is a linear map. $\underline h^{-1}$ maps a vector $X$ to its gauge invariant counterpart $\underline h^{-1}(X)$. The adjoint map $\overline h$ maps a derivative $\partial f$ to its gauge invariant counterpart $\overline h(\partial f)$.

Only operations on gauge-invariant quantities are, in general, geometrically meaningful. So only $\overline h(\partial f) \cdot \underline h^{-1}(X)$ is a meaningful quantity. But the definition of the adjoint means that $\overline h(A) \cdot \underline h^{-1}(B) = \underline h \underline h^{-1}(A) \cdot B = A \cdot B$ for any $A$ and $B$.

Thus, the dependence of the dot product on the gauge field cancels out, but only when one quantity depends on $\underline h^{-1}$ (that is, it's a vector) and the other depends on $\overline h$ (that is, it's a covector). This is a time-honored shortcut to getting geometrically meaningful results without having to pay attention to the gauge. It says that, for instance, the dot product of a basis vector and its corresponding basis covector (say $e_i \cdot e^i$, no summation implied) should always be $1$, regardless of gauge.

In other words, if you started with everything being vectors and then defined covectors, you would pick up a factor of the metric $\underline g = \overline h^{-1} \underline h^{-1}$ to do so, but bringing that covector back to being a vector just picks up $\overline g^{-1}$, so the metric cancels itself out.


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