# Help operating with differentials in Geometric Algebra

I'm trying to learn General Relativity with Geometric Algebra by following the article Spacetime Geometry with Geometric Calculus, but I'm finding some algebraic problems when dealing with expressions like the following, eq. 96:

$$\partial_a \cdot (\dot{D} \wedge \dot{R}(a \wedge b))= \dot{R}(\dot{D} \wedge b) - D \wedge R(b) = 0$$

$$D$$ being the covariant derivative operator and $$R(a \wedge b)$$ being the Riemann tensor.

How does the $$D$$ operator enter into the argument of the Riemann tensor? And why is there a minus sign for the Ricci tensor term?

I have read through Clifford Algebra to Geometric Calculus and Geometric Algebra for physicists but I haven't found proper explanations about how these operations are performed.

• This is contracting the Bianchi identity, which is covered (although it's hard to follow) in Clifford Algebra to Geometric Calculus section 5-1. See equations 5-1.13a and 5-1.22 . Commented Apr 9, 2022 at 3:49

You're asking two questions about how to contract the Bianchi identity using geometric algebra. The question about where the minus sign comes from is an issue with how to distribute product operators in geometric algebra. The question about how $$D$$ winds up inside of $$R$$'s argument is an issue with a specific application of geometric calculus. I'll address the two issues in general first, and then apply the results to this specific problem.

First, the geometric algebra issue:

It's surprisingly hard to find the identities for how inner and outer products distribute over other products, when not all of the multivectors involved are vectors. Clifford Algebra to Geometric Calculus does have these identities, though, on page 12. We'll only need one of these identities, but I'll just list them all here to make them easier to find for other people who may have questions about how to distribute among the various products in geometric algebra.

For $$a$$ a vector, $$A_r$$ a homogeneous multivector of grade $$r$$ and $$B$$ a multivector, the inner and outer products distribute over the geometric product as

\begin{aligned} a\cdot(A_r B)&=a\cdot A_r B+(-1)^r A_r a\cdot B\\ &=a\wedge A_r B-(-1)^r A_r a\wedge B \end{aligned}

\begin{aligned} a\wedge(A_r B)&=a\wedge A_r B-(-1)^r A_r a\cdot B\\ &=a\cdot A_r B+(-1)^r A_r a\wedge B\ . \end{aligned}

For $$B_s$$ a homogeneous multivector, we can perform grade projection on the above identities to get

$$a\cdot(A_r\wedge B_s)=(a\cdot A_r)\wedge B_s+(-1)^r A_r \wedge(a\cdot B_s)$$

$$a\wedge(A_r\cdot B_s)=(a\cdot A_r)\cdot B_s+(-1)^r A_r \cdot(a\wedge B_s)\ .$$

Next, the geometric calculus issue:

First, note that both the vector derivative and the covariant derivative are vector operators, which can be treated algebraically just like any other vector. This is perhaps easiest to see if you express things in terms of components. Split a vector $$a$$ into components in the $$\{e_i\}$$ frame as $$a=a^i e_i$$. For the vector derivative, if a term contains $$\dot{\partial}_a$$ somewhere and $$\dot{F}(a)$$ somewhere, you can replace the $$\dot{\partial}_a$$ with the vector $$e^i$$, and replace $$\dot{F}(a)$$ with $$\frac{\partial}{\partial a^i}F(a)$$.

Something similar still holds with the covariant derivative, even though the covariant derivative involves a projection onto the tangent space. With $$\{e_i\}$$ now a basis for the tangent space, if a term includes $$\dot{D}$$ somewhere and $$\dot{F}$$ somewhere, where $$D$$ is the covariant derivative, you can replace $$\dot{D}$$ with $$e^i$$ and $$\dot{F}$$ with $$(e_i \cdot D)F$$. Note that $$(e_i\cdot D)$$ is a scalar operator, so it commutes with everything and can go anywhere as a factor within the term, although overdots are needed if the $$(e_i\cdot D)$$ isn't placed right before the $$F$$.

If a function $$F(a)$$ is linear in the vector $$a=a^i e_i$$, we have that $$\frac{\partial}{\partial a^i}F(a)=\lim_{\epsilon\to 0}\frac{F(a+\epsilon e_i)-F(a)}{\epsilon}=\lim_{\epsilon\to 0}\frac{F(\epsilon e_i)}{\epsilon}=F(e_i)\ .$$ This means that for any vector $$v$$, \begin{aligned} (\dot{\partial}_a\cdot v)\dot{F}(a)&=(e^i\cdot v)\frac{\partial}{\partial a^i}F(a)\\ &=v^i\frac{\partial}{\partial a^i}F(a)\\ &=v^i F(e_i)\\ &=F(v^i e_i)\\ &=F(v)\ , \end{aligned} where in the next to last line we again relied on $$F$$ being linear.

We're now prepared to directly address the problem in question. We have \begin{aligned} 0&=\partial_a\cdot(\dot{D}\wedge\dot{R}(a\wedge b))\\ &=(\partial_a\cdot \dot{D})\dot{R}(a\wedge b))-\dot{D} \wedge(\partial_a\cdot R(a\wedge b))\\ &=\dot{R}(\dot{D}\wedge b)-\dot{D} \wedge(\partial_a\cdot R(a\wedge b))\\ &=\dot{R}(\dot{D}\wedge b)-\dot{D} \wedge R(b)\ . \end{aligned}

In this equation,

Line 1 is contracting $$\partial_a$$ with the Bianchi identity.

Line 2 uses the identity above for distributing the inner product over the outer product, with $$a\to\partial_a$$, $$A_r\to D$$ and $$B_s\to R(a\wedge b)$$. Note that $$r=1$$ and $$s=2$$. The wedge product goes away because the dot product just results in a scalar, and $$1\wedge B_s=B_s$$.

Line 3 uses the calculus result we derived, with $$\partial_a\to \partial_a$$, $$v\to D$$ and $$F(a)\to R(a+b)$$.

Line 4 uses the definition of the Ricci tensor, equation 83 in the paper.