In geometric algebra we have the derivative by a vector of a vector field $$\nabla V=\nabla \cdot V+\nabla \wedge V$$ While in tensor analysis we have $$\nabla V=\frac{1}{n}I(n)\nabla \cdot V +\nabla \wedge V+\sigma (V)$$ where I(n) is the n-dimensional identity matrix and $\sigma (V)$ is the symmetric traceless part of the matrix associated with volume-preserving shear.

Where do we find the missing shear term in GA?

  • $\begingroup$ What is $\nabla \wedge V$? $\endgroup$
    – Kaster
    Commented Sep 30, 2013 at 22:12
  • $\begingroup$ In the GA equation, it is the curl bivector. In the matrix equation, it is the anti-symmetric part. $\endgroup$ Commented Oct 1, 2013 at 5:43
  • $\begingroup$ Could you please refer to some literature about your $\nabla V$ in tensor analysis? I've never seen that. $\nabla$ in differential geometry is covariant derivative operator. $\endgroup$
    – Shuchang
    Commented Oct 1, 2013 at 13:15
  • $\begingroup$ here $\endgroup$ Commented Oct 1, 2013 at 14:18

1 Answer 1


It helps to write the tensor notion of the derivative in the GA/GC language. The tensor idea of the derivative is more like this:

$$\underline V(a) = (a \cdot \nabla) V$$

where $\underline V$ is a linear operator on the space.

Since a linear operator can be decomposed into a traceless symmetric part, an antisymmetric part, and a traceful part, we can write $\underline V$ as

$$\underline V(a) = \frac{1}{n} T \underline I(a) + \underline A(a) + \underline S(a)$$

where $T$ is the trace, $\underline I$ is the identity, $\underline A$ is antisymmetric, and $\underline S$ is symmetric and tracefree.

In GA/GC, operations like the trace of a linear operator are performed not by summing through bases but with the machinery of geometric calculus and differentiation. The trace is

$$\partial_a \cdot \underline V(a) = T$$

where $\partial_a$ is vector differentiation with respect to $a$ (so that $\nabla = \partial_x$ when $x$ is the position vector).

Just as the trace is a scalar invariant, there is a bivector invariant:

$$\partial_a \wedge \underline V(a) = \partial_a \wedge\underline A(a) \equiv A$$

(This curl produces zero on the identity and the symmetric-tracefree part. In fact, this is one way to define what it means for an operator to be symmetric.)

The geometric calculus vector derivative is then defined as the geometric product of $\partial_a$ with $a \cdot \nabla V$:

$$\nabla V = \partial_a ([a \cdot \nabla] V) = T + A$$

So we identify the trace with the divergence and $A$ with the curl.

But, as you astutely noted, the symmetric tracefree part $\underline S$ appears in neither of these expressions. It is still absolutely there in the definition of $\underline V(a)$, and perhaps you can say geometrically that these terms produce bivectors of equal magnitude but opposite orientation, so they cancel. To tell the truth, the geometric significance of the symmetric tracefree part is still somewhat mysterious to me; I know Alan MacDonald has said similar things (that GA/GC does not handle symmetric tracefree tensors with a useful geometric interpretation as yet).

Nevertheless, the GA/GC viewpoint on linear algebra is a fascinating one. It's still one of my personal holy grails to understand the GA/GC version of the Cayley-Hamilton theorem for trying to find eigenblades of linear operators. I think this is particularly important when maps are not operators, too, for instance with the moment of inertia tensor, which naturally maps bivectors to bivectors and thus is resistant to the usual approach of finding a characteristic polynomial (or it should be).

  • $\begingroup$ any reference which shows how a linear operator can be decomposed into a traceful part, a traceless symmetric and an antisymmetric part ? $\endgroup$
    – user297514
    Commented Jan 9, 2017 at 11:12
  • $\begingroup$ @gansub Given an operator $T$, you can decompose it into $T=T^+ + T^-= (T+T^T)/2+(T-T^T)/2$ $T^T$ is the standard transpose. $T^+$ can then be broken up into $T^+= (T^+ -tr(T)I/n)+ tr(T)I/n$. tr is the trace and $I$ is the identity matrix and $n$ is the dimension. It’s pretty easy to see why $T_0= T^+ - tr(T)I/n$ is tracefree as the trace is linear. Substituting the middle equation into the first, you get $T= tr(T)I /n+ T^- + T_0$. $\endgroup$
    – Lave Cave
    Commented Jul 19, 2022 at 4:54

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