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## Covariant derivatives and parallelism for tensors November 3, 2009

Posted by Akhil Mathew in differential geometry, MaBloWriMo.
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Time to continue the story for covariant derivatives and parallelism, and do what I promised yesterday on tensors.

Fix a smooth manifold ${M}$ with a connection ${\nabla}$. Then parallel translation along a curve ${c}$ beginning at ${p}$ and ending at ${q}$ leads to an isomorphism ${\tau_{pq}: T_p(M) \rightarrow T_q(M)}$, which depends smoothly on ${p,q}$. For any ${r,s}$, we get isomorphisms ${\tau^{r,s}_{pq} :T_p(M)^{\otimes r} \otimes T_p(M)^{\vee \otimes s} \rightarrow T_q(M)^{\otimes r} \otimes T_q(M)^{\vee \otimes s} }$ depending smoothly on ${p,q}$. (Of course, given an isomorphism ${f: M \rightarrow N}$ of vector spaces, there is an isomorphism ${M^* \rightarrow N^*}$ sending ${g \rightarrow g \circ f^{-1}}$—the important thing is the inverse.)

In the future, I will often omit the superscript ${r,s}$ because it will cause no confusion.

Thus we have described parallel translation on (potentially mixed) tensors. There are a few things I claim:

1. For ${v,w}$, we have ${\tau_{pq}(v \otimes w) = \tau_{pq} v \otimes \tau_{pq} w}$.
2. ${\tau_{pq}}$ commutes with contractions.
3. ${\tau_{pq}}$ commutes with the symmetrization and alternation maps on unmixed (i.e. fully covariant or contravariant) tensors.

The first follows from the description, and so does the second from the parenthetical remark. The third is clear. In particular, the third implies that we can talk about parallel translation of a Riemannian metric or a form.

Now, as in the previous post, it is possible to define a covariant derivative on these tensors. Let ${Y \in T_p(M)}$ and let ${X}$ be a tensor field. Choose a curve ${c}$ tangent to ${Y}$ at ${0}$. Define the covariant derivative $\displaystyle \nabla_Y(X) := \lim_{s \rightarrow 0} \frac{ \tau_{p, c(s)}^{-1} X(c(s)) - X(p) }{s}.$

For the moment, let’s ignore the question of whether this is well-defined, i.e. that the choice of ${c}$ doesn’t matter. This will follow from the arguments below.

I claim that

1. ${\nabla_Y( A \otimes B) = \nabla_Y A \otimes B + A \otimes \nabla_Y B}$
2. ${\nabla_Y}$ commutes with contractions.

The first is basically the product rule, and follows from the first remark on parallel translation, because ${\nabla_Y(A \otimes B)}$ can be split into two parts: $\displaystyle \lim_{s \rightarrow 0} \frac{ \tau_{p, c(s)}^{-1} A (c(s)) \otimes \tau_{p, c(s)}^{-1}B (c(s)) -\tau_{p, c(s)}^{-1} A (c(s)) \otimes B (p) }{s}$

and $\displaystyle \lim_{s \rightarrow 0} \frac{ \tau_{p, c(s)}^{-1} A (c(s)) \otimes B (p) - A(p) \otimes B(p) }{s}.$

The second follows from the corresponding of parallel translation. These two together show that the choice of the curve ${c}$ doesn’t matter for any tensor, since it doesn’t matter for zero-tensors (i.e. functions) or vector fields.

Finally, from all this we can also talk about the covariant derivative of a tensor field along a curve, using, e.g., the limit. It is the same story though, and I already fear that I have spent too much time on this.

Now we will specialize to the case of a Riemannian metric. Say that a metric ${(.,.)}$ and a connection ${\nabla}$ are compatible if $\displaystyle \nabla_Y( (.,.)) = 0, \forall Y.$

This will be an important notion in the future.

I claim that this is equivalent to saying that parallel translation preserves inner products on tangent vectors. Indeed, assume the bolded statement, and fix a curve ${c}$ and parallel vector fields ${X_1, X_2}$ on ${c}$. Let ${c(0)=p, c'(0)=Y}$. Then ( ${D}$ denoting the covariant derivative along a curve) $\displaystyle D( (X_1, X_2) )(0) = Yg(X_1,X_2)(0) = 0.$

We can extend ${X_1,X_2}$ locally to vector fields on an open set and so $\displaystyle 0 = \nabla_Y( C( (.,.) \otimes (X_1 \otimes X_2 )) )$

where ${C}$ is contraction, so $\displaystyle 0 = C( \nabla_Y( (.,.)) \otimes (X_1)_p \otimes (X_2)_p ) ,$

which means that the 2-tensor ${\nabla_Y( (.,.))}$ vanishes on the pair ${(X_1)_p, (X_2)_p}$. Since ${Y, X_1, X_2}$ were arbitrary, this completes one half the proof. This argument can be reversed, which gives the other half.

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