# What's the intuition behind the shear not being possible to be obtained with exterior differentiation?

Allow me to present some context, first. If we have a $$K$$-vector space $$E$$ and a non-degenerate symmetric bilinear form (a pseudo-riemannian metric) $$g$$, then we can project any 2nd order tensor (the exact covariance and contravariance doesn't really matter because we have a metric) into three subspaces of the space of tensors which are invariant under automorphisms of $$E$$ (i.e. changes of basis). These projections represent its trace, its antisymmetric component (I'm considering $$\text{char}(K)\neq2$$, of course) andits traceless symmetric component. For a generic tensor, this decomposition looks like this:

$$T_{ij}= \frac{1}{3}T^k{}_k\space g_{ij}+T_{[ij]}+\left(T_{(ij)} - \frac{1}{3}T^k{}_k\space g_{ij}\right).$$

If we do this to the covariant derivative of a 1st order tensor field $$\vec{v}\in T_P\mathbb{R}^3$$, we find that

$$v_{i;j} = \nabla_{j\space}v_i = \frac{1}{3}g_{ij}(\text{div}\space \vec{v}) + \frac{1}{2}\varepsilon_{ijk}(\text{curl}\space\vec{v})^k + \frac{1}{2}(\text{shear}\space\vec{v})_{ij}.$$

Now, we can also find the divergence and the curl of a 1st order tensor field using the exterior derivative of this manifold and the Hodge star thingie the metric grants us:

$$\text{div}=\star\space\text{d}\space\star,\space\space\space\space\space\text{curl}=\star\space\text{d}$$

This makes sense because the divergence is a scalar field and the curl is a (pseudo)vector field, and both are differential forms (antisymmetric covariant tensor fields) when they have their indices lowered. The shear, though, is a symmetric 2nd order tensor field, so it can't be a differential 2-form.

My questions are:

• Is everything I just said true?
• If so, why are the flux and circulation of a vector field around a point related to differential forms, but not its shear? Does it have something to do with the fact that the shear isn't usually taught in undergrad physics, math and engineering courses (besides the fact that, you know, it's a 2nd order tensor and nobody wants to deal with those if it's not necessary).