Is this identity about the Ricci curvature true?


Let $\M$ be a Riemannian manifold, $\nabla^{T\M}$ is its Levi-Civita connection. Given $X \in \Gamma(T\M)$, we consider $\nabla^{T\M} X$ as a linear map (vector bundle morphism) $T\M \to T\M$.

Is it true that for all $X,Y \in \Gamma(T\M)$ $$\Ric(Y,X)=\tr (\nabla^{T\M}Y) \cdot \tr(\nabla^{T\M}X) -\tr ( \nabla^{T\M}Y \circ \nabla^{T\M}X), \tag{1}$$

where $\Ric$ is the Ricci curvature of $\M$.

Motivation: I somehow derived (in a very cumbersome way) the equality $$\int_{\M} \Ric(Y,X)= \int_{\M} \tr (\nabla^{T\M}Y) \cdot \tr(\nabla^{T\M}X) -\tr ( \nabla^{T\M}Y \circ \nabla^{T\M}X) , \tag{2}$$

(I assume $\M$ is closed and oriented; the integration is w.r.t the Riemannian volume form of $\M$).

Edit:

As observed by levap, equality $(1)$ is false. Here is a proof of equality $(2)$, based on a few lemmas in my answer:

First note that $\tr (\nabla^{T\M}Y)=\div Y,\tr (\nabla^{T\M}X)=\div X$, so equality $(2)$ is nothing but

$$\int_{\M} \Ric(Y,X)= \int_{\M} \div Y\cdot\div X -\tr ( \nabla^{T\M}Y \circ \nabla^{T\M}X) . \tag{3}$$

We now turn to analyse both summands of the RHS, up to divergence terms: (since the integral of a divergence is zero, this does not matter for the integral):

The first observation (by Anthony Carapetis) is that $$(\mathrm{tr}_{13} \nabla^2 Y)(X) = \mathrm{div}(\nabla_X Y) - \mathrm{tr}(\nabla Y \circ \nabla X). \tag{4}$$

(For a proof, see lemma 2 in my answer).

For the other term, we use the identity $\div(fX)=f\div X+\tr(df \otimes X)$, for $f=\div Y$:

$$\div(\div Y X)=\div Y\div X+\tr\big(d\div Y \otimes X\big). \tag{5}$$ Combining equations $(4),(5)$ we obtain that equality $(3)$ is equivalent to

$$\int_{\M} \Ric(Y,X)= \int_{\M} -\tr\big(d\div Y \otimes X\big) +(\mathrm{tr}_{13} \nabla^2 Y)(X). \tag{6}$$

Now, we use $$\tr\big(d\div Y \otimes X\big) = (\tr_{23} \nabla^2 Y)(X), \tag{7}$$ (See lemma 3 in my answer)

which implies $(6)$ is equivalent to

$$\int_{\M} \Ric(Y,X)= \int_{\M} (\mathrm{tr}_{13} \nabla^2 Y-\tr_{23} \nabla^2 Y)(X). \tag{8}$$

But, lemma 1 in my answer implies that now the integrands are equal: $$\Ric(Y,X)= (\mathrm{tr}_{13} \nabla^2 Y-\tr_{23} \nabla^2 Y)(X).$$

To summarize, we showed the original integrands are the same up to divergence terms so their integrals are equal.

• The integral version is what you get by writing out the Ricci curvature in terms of second covariant derivatives and then integrating by parts in the most obvious way. – Anthony Carapetis Dec 10 '17 at 14:09
• In abstract indices it's simply $\mathrm {Rc}(X,Y) = X^j (\nabla_i \nabla_j Y^i - \nabla_j \nabla_i Y^i).$ Integrating by parts moves the outer derivatives on to $X$. It can be cumbersome to handle this kind of thing with index-free notation: the first term is something like like $$(\mathrm{tr}_{13} \nabla^2 Y)(X) = \mathrm{div}((\nabla Y)(X)) - \mathrm{tr}(\nabla Y \circ \nabla X),$$ and the integral of a divergence is zero. – Anthony Carapetis Dec 11 '17 at 12:35
• Thanks, I agree with your last formula, but I don't see how to handle similarly the first term $\tr (\nabla^{T\M}Y) \cdot \tr(\nabla^{T\M}X)=\div X \div Y$ in a useful way (as you can see in my edit, I tried something, but I am not sure it's helpful). Are you convinced that using the abstract indices gives a quick way to see this? (I really should learn how to use them properly). – Asaf Shachar Dec 11 '17 at 17:32
• You're basically there, though I think you've dropped a sign somewhere. The integrands on both sides of $(6)$ are equal (after correcting the sign). Just check that $\tr\big(d\div Y \otimes X\big) = (\tr_{23} \nabla^2 Y)(X),$ and then that $\Ric(Y,X) = (\tr_{23} \nabla^2 Y - \tr_{13} \nabla^2 Y)(X),$ which follows from the fact that $X,Z\mapsto R(X,Z)Y$ is twice the antisymmetric part of $\nabla^2 Y.$ – Anthony Carapetis Dec 12 '17 at 2:05
• And yes, it's sometimes very helpful to do calculations in abstract index notation. In my experience the hardest part about it is convincing yourself it's not cheating - here the whole calculation is simply $$\int \Ric(X,Y) = \int X^j \nabla_i \nabla_j Y^i - X^j \nabla_j \nabla_i Y^i = \int - \nabla_i X^j \nabla_j Y^i + \nabla_j X^j \nabla_i Y^i,$$ which seems far too easy when compared to your writeup. – Anthony Carapetis Dec 12 '17 at 10:06

The right hand side is not tensorial in $X,Y$ so you definitely don't have a trivial equality. Consider for example the case $M = \mathbb{R}^n$ with the flat metric. The left hand side is zero but the right hand side is

$$\operatorname{tr}(dX) \operatorname{tr}(dY) - \operatorname{tr}(dX \circ dY)$$

where we identify the vector fields $X,Y$ on $\mathbb{R}^n$ with smooth maps $X,Y \colon \mathbb{R}^n \rightarrow \mathbb{R}^n$ and then the full covariant derivatives $\nabla X, \nabla Y$ are identified with the differentials $dX, dY$. When $n > 1$, the identity

$$0 = \operatorname{tr}(dX) \operatorname{tr}(dY) - \operatorname{tr}(dX \circ dY)$$

is false already for linear vector fields $X,Y$.


Lemma 1:

$$\Ric(X,Y) = (\tr_{13} \nabla^2 Y - \tr_{23} \nabla^2 Y)(X)$$

Proof:

We first note that $$\nabla^2 Y(X,Z)-\nabla^2 Y(Z,X)=(\nabla_X \nabla_Z Y-\nabla_{\nabla_X Z}Y)-(\nabla_Z \nabla_X Y-\nabla_{\nabla_Z X}Y)=$$

$$\nabla_X \nabla_Z Y-\nabla_Z \nabla_X Y-\nabla_{(\nabla_X Z-\nabla_Z X)}Y=\nabla_X \nabla_Z Y-\nabla_Z \nabla_X Y-\nabla_{[X,Z]}Y=R(X,Z)Y.$$

Now,

$$(\tr_{23} \nabla^2 Y - \tr_{13} \nabla^2 Y)(X)=\sum_i \langle \nabla^2 Y(X,e_i)-\nabla^2 Y(e_i,X),e_i \rangle=\sum_i \langle R(X,e_i)Y,e_i \rangle=$$

$$-\sum_i \langle R(e_i,X)Y,e_i \rangle=-\Ric (X,Y).$$

Comment: Here is why $$\nabla^2 Y(X,Z)=\nabla_X \nabla_Z Y-\nabla_{\nabla_X Z}Y :\tag{1}$$

By definition, we have $$\nabla^2 Y(X,Z):=(\nabla^{T^*M \otimes TM}_X \nabla Y)(Z).$$

$$\nabla^{TM}_X \big((\nabla^{TM} Y)(Z))=(\nabla^{T^*M \otimes TM}_X \nabla Y)(Z)+(\nabla^{TM} Y)(\nabla_X^{TM} Z),$$

which is equivalent to

$$\nabla^{TM}_X \nabla_Z^{TM} Y= \nabla^2 Y(X,Z)+\nabla_{\nabla_X^{TM} Z}^{TM}Y.$$

Lemma 2: $$(\mathrm{tr}_{13} \nabla^2 Y)(X) = \mathrm{div}(\nabla_X Y) - \mathrm{tr}(\nabla Y \circ \nabla X).$$

Proof:

Using equation $$(1)$$, we get

$$(\mathrm{tr}_{13} \nabla^2 Y)(X)=\sum_i \langle \nabla^2 Y(e_i,X),e_i \rangle=\sum_i \langle \nabla_{e_i} \nabla_X Y-\nabla_{\nabla_{e_i} X}Y,e_i \rangle=$$

$$\sum_i \langle \nabla_{e_i} (\nabla_X Y),e_i \rangle-\sum_i \langle \nabla_{\nabla_{e_i} X}Y,e_i \rangle=$$

$$\mathrm{div}(\nabla_X Y)-\sum_i \langle \nabla Y(\nabla_{e_i} X),e_i \rangle=$$

$$\mathrm{div}(\nabla_X Y)-\sum_i \langle \nabla Y(\nabla X(e_i)),e_i \rangle=$$

$$\mathrm{div}(\nabla_X Y)-\sum_i \langle (\nabla Y \circ \nabla X)(e_i),e_i \rangle=\mathrm{div}(\nabla_X Y) - \mathrm{tr}(\nabla Y \circ \nabla X).$$


Lemma 3:

$$\tr\big(d\div Y \otimes X\big) = (\tr_{23} \nabla^2 Y)(X).$$

Proof:

Let $$p \in M$$. Let $$e_i$$ be a frame induced by normal coordinates centered at $$p$$. In particular, $$e_i(p)$$ is an orthonormal basis for $$T_pM$$, and $$\nabla_{e_j} {e_i}=0$$.

On the one hand

$$A:=(\tr_{23} \nabla^2 Y)(X)= \sum_i \langle \nabla^2 Y(X,e_i),e_i \rangle=\sum_i \langle \nabla_X \nabla_{e_i} Y-\nabla_{\nabla_X {e_i}}Y,e_i \rangle= \tag{1*}$$

$$\sum_{ij} X_j \langle \nabla_{e_j} \nabla_{e_i} Y-\nabla_{\nabla_{e_j} {e_i}}Y,e_i \rangle=\sum_{ij} X_j \langle \nabla_{e_j} \nabla_{e_i} Y,e_i \rangle=\sum_j X_ja_j,$$

where

$$a_j=\sum_i \langle \nabla_{e_j} \nabla_{e_i} Y,e_i \rangle \tag{2*}$$

On the other hand,

$$B:=\tr\big(d\div Y \otimes X\big)=\sum_j \langle (d\div Y)(e_j)X,e_j \rangle=\sum_j \langle b_jX,e_j \rangle=\sum_j b_jX_j, \tag{3*}$$

where

$$b_j:=(d\div Y)(e_j)=e_j \cdot \div Y=e_j \cdot \sum_i \langle \nabla_{e_i}Y,e_i \rangle=\sum_i e_j \cdot \langle \nabla_{e_i}Y,e_i \rangle \tag{4*}=$$

$$\sum_i \big( \langle \nabla_{e_j}\nabla_{e_i}Y,e_i \rangle+ \langle \nabla_{e_i}Y,\nabla_{e_j} e_i \rangle \big)=\sum_i \langle \nabla_{e_j}\nabla_{e_i}Y,e_i \rangle=a_j.$$