## Derivations and the curvature tensor

Let ${M}$ be a Riemannian manifold with Riemannian connection ${\nabla}$. A connection is a thing that knows how to differentiate a vector field ${Y}$ in the direction of a vector field ${X}$; the result is denoted by ${\nabla_X Y}$ and is also a vector field. For consistency of notation, it is convenient to write ${\nabla_X f }$ for the derivative of scalar function ${f}$ in the direction ${X}$, even though this derivative does not need a connection: vector fields are born with the ability to differentiate functions.

The pairs ${(f,Y)}$, with ${f}$ a scalar function and ${Y}$ a vector field, form a funky nonassociative algebra ${\mathcal{A}}$ described in the previous post. And ${\nabla_X}$ is a derivation on this algebra, because

• ${\nabla_X(fY) = f\nabla_X Y + (\nabla_X f)Y }$ by the definition of a connection
• ${\nabla_X\langle Y, Z\rangle = \langle \nabla_X Y, Z\rangle + \langle Y, \nabla_X Z\rangle}$ by the metric property of the Riemannian connection.

Recall that the commutator of two derivations is a derivation. Or just check this again: if ${{}'}$ and ${{}^\dag}$ are derivations, then

${ {(ab)'}^\dag = (a'b+ab')^\dag = {a'}^\dag b+a'b^\dag +a^\dag b'+a{b'}^\dag }$
${ {(ab)^\dag}' = (a^\dag b+ab^\dag)' = {a^\dag }' b+a^\dag b' +a' b^\dag +a{b^\dag}' }$

and the difference ${{(ab)'}^\dag-{(ab)^\dag}'}$ simplifies to what it should be.

Thus, for any pair ${X,Y}$ of vector fields the commutator ${\nabla_X \nabla_Y-\nabla_Y\nabla_X}$ is a derivation on ${\mathcal{A}}$. The torsion-free property of the connection tells us how it works on functions:

$\displaystyle (\nabla_X \nabla_Y-\nabla_Y\nabla_X) f = \nabla_{[X,Y]}f=\nabla_{\nabla_XY}f -\nabla_{\nabla_YX}f$

Subtracting ${\nabla_{[X,Y]}f}$ from the commutator, we get a derivation that kills scalar functions,

$\displaystyle R(X,Y) = \nabla_X \nabla_Y-\nabla_Y\nabla_X - \nabla_{[X,Y]}$

But a derivation that kills scalar functions is linear over functions:

$\displaystyle R(X,Y)(fZ) = R(X,Y)(f)\, Z + f\,R(X,Y)Z = f\,R(X,Y)Z$

In plain terms, ${R(X,Y)}$ processes any given vector field ${Z}$ pointwise, applying some linear operator ${L_p}$ to the vector ${Z_p}$ at every point ${p}$ of the manifold. No derivatives of ${Z}$ are actually taken, either of first or of second order.

Moreover, the derivation property immediately implies that ${R(X,Y)}$ is a skew-symmetric operator: for any vector fields ${Z,W}$

$\displaystyle \langle R(X,Y)Z,W\rangle + \langle R(X,Y)W,Z\rangle = R(X,Y)\langle Z,W\rangle =0$

because ${R(X,Y)}$ kills scalar functions.

The other kind of skew-symmetry was evident from the beginning: ${R(X,Y)=-R(Y,X)}$ by definition.

What is not yet evident is that ${R(X,Y)}$ is also a tensor in ${X}$ and ${Y}$, that is, it does not differentiate the direction fields themselves. To prove this, write ${R(X,Y)=\nabla_{X,Y}^2-\nabla_{Y,X}^2}$ where $\displaystyle \nabla_{X,Y}^2 = \nabla_X \nabla_Y - \nabla_{\nabla_X Y}$ should be thought of as the pointwise second-order derivative in the directions ${X,Y}$ (i.e., the result of plugging two direction vectors into the Hessian matrix). By symmetry, it suffices to show that ${\nabla_{X,Y}^2}$ is a tensor in ${X}$ and ${Y}$. For ${X}$, this is clear from the definition of connection. Concerning ${Y}$, we have

${ \nabla_{X,fY}^2 = \nabla_X (f\nabla_{Y}) - \nabla_{(\nabla_X f) Y+f\nabla_X Y} }$
${= f \nabla_X \nabla_{Y} + (\nabla_X f )\nabla_{Y} - (\nabla_X f) \nabla_{ Y} - f \nabla_{\nabla_X Y} }$
${= f \nabla_{X,Y}^2 }$

That’s it, we have a tensor that takes three vector fields ${X,Y,Z}$ and produces another one, denoted ${R(X,Y)Z}$. Now I wonder if there is a way to use the language of derivations to give a slick proof of the first Bianchi identity, ${R(X,Y)Z+R(Y,Z)X+R(Z,X)Y=0}$

To avoid having two picture-less posts in a row, here is something completely unrelated:

This is the image of the unit circle ${|z|=1}$ under the polynomial ${z^3-\sqrt{3}\,\bar z}$. Which area is larger: red or green? Answer hidden below.

They are equal.

## Derivations

A map ${D}$ is a derivation if it satisfies the Leibniz rule: ${D(ab)=D(a)b+aD(b)}$. To make sense out of this, we need to be able to

• multiply arguments of ${D}$ together
• multiply values of ${D}$ by arguments of ${D}$
• add the results

For example, if ${D\colon R\to M}$ where ${R}$ is a ring and ${M}$ is a two-sided module over ${R}$, then all of the above makes sense. In practice it often happens that ${M=R}$. In this case, the commutator (Lie bracket) of two derivations ${D_1,D_2}$ is defined as ${[D_1,D_2]=D_1\circ D_2-D_2\circ D_1}$ and turns out to be a derivation as well. If ${R}$ is also an algebra over a field ${K}$, then ${K}$-linearity of ${D}$ can be added to the requirements of being a derivation, but I am not really concerned about that.

What I am concerned about is that two of my favorite instances of the Leibniz rule are not explicitly covered by the ring-to-module derivations. Namely, for smooth functions ${\varphi\colon{\mathbb R}\rightarrow{\mathbb R}}$, ${F\colon{\mathbb R}\rightarrow{\mathbb R}^n}$ and ${G\colon{\mathbb R}\rightarrow{\mathbb R}^n}$ we have

$\displaystyle (\varphi F)' = \varphi' F + \varphi F' \quad \text{and} \quad (F\cdot G)' = F'\cdot G+F\cdot G' \ \ \ \ \ (1)$

Of course, ${{\mathbb R}^n}$ could be any ${{\mathbb R}}$-vector space ${V}$ with an inner product.

It seems that the most economical way to fit (1) into the algebraic concept of derivation is to equip the vector space ${{\mathbb R}\oplus V}$ with the product

$\displaystyle (\alpha,u)(\beta,v)= (\alpha\beta+u\cdot v, \alpha v+\beta u) \ \ \ \ \ (2)$

making it a commutative algebra over ${{\mathbb R}}$. Something tells me to put ${-u\cdot v}$ there, but I resist. Actually, I should have said “commutative nonassociative algebra”:

$\displaystyle \{(\alpha,u)(\beta,v)\}(\gamma,w) = (\alpha\beta+u\cdot v, \alpha v+\beta u) (\gamma,w) \\ \\ = (\alpha\beta\gamma+\gamma u\cdot v+ \alpha v\cdot w+\beta u\cdot w, \alpha\beta w + \gamma \alpha v+ \gamma \beta u +(u\cdot v) w)$

Everything looks nice, except for the last term ${(u\cdot v) w}$, which destroys associativity.

Now we can consider maps ${{\mathbb R}\rightarrow {\mathbb R}\oplus V}$, which are formal pairs of scalar functions and vector-valued functions. The derivative acts component-wise ${(\varphi,F)'=(\varphi',F')}$ and according to (1), it is indeed a derivation:

$\displaystyle \left\{(\varphi,F)(\psi,G)\right\}'= (\varphi,F)'(\psi,G)+(\varphi,F)(\psi,G)' \ \ \ \ \ (3)$

Both parts of (1) are included in (3) as special cases ${(\varphi,0)(0,F)}$ and ${(0,F)(0,G)}$.

If (2) has a name, I do not know it. Clifford algebras do a similar thing and are associative, but they are also larger. If I just want to say that (1) is a particular instance of a derivation on an algebra, (2) looks like the right algebra structure to use (maybe with ${-u\cdot v}$ if you insist). If ${V}$ has no inner product, the identity ${(\varphi F)' = \varphi' F + \varphi F'}$ can still be expressed via (2) using the trivial inner product ${u\cdot v=0}$.