# Second-order tensors as linear operators

A tensor is formally defined as an object whose components obey some transformation rules. I, however, find it more intuitive to look at (second-order) tensors as a linear operator/function between two vectors. Thus the stress tensor $$\sigma$$ connects the surface normal $$\vec{n}$$ to the force acting on that surface as $$\vec{f} = \sigma \vec{n}$$. Similarly, the moment of inertia tensor connects the angular momentum to the angular velocity, so $$\vec{L} = I \vec{\omega}$$.

There are, however, two tensors I never saw in this context. One of them is the strain tensor, $$\epsilon_{ij}$$, the other one is the metric tensor $$g_{ij}$$ of a 2D surface embedded in the 3D space. What would be the two vectors the strain tensor connects such that $$\vec{a} = \epsilon \vec{b}$$? Also, what would be the two vectors the metric tensor connects?

## 1 Answer

First some review, to establish context. Second order tensors come in three types: $$(0, 2)$$, $$(1, 1)$$, and $$(2, 0)$$, depending on the number of upper and lower indices. In general, an $$(n, m)$$ tensor is a multilinear real-valued function of $$n$$ one-forms and $$m$$ vectors. However, by currying the function, an $$(n, m)$$ tensor can also be equivalently viewed as a multilinear map of $$n-k$$ one-forms and $$m-l$$ vectors to $$l$$ one-forms and $$k$$ vectors. Your interpretation of a $$(1, 1)$$ tensor is of course the special case $$n = m = k = 1$$, $$l = 0$$.

The metric is naturally a $$(0, 2)$$ tensor, so we can think about it two ways according to the above:

• $$l = 0$$, where the metric is a real-valued multilinear map on two vectors, i.e. precisely the inner product of those vectors
• $$l = 1$$, where the metric maps vectors to one-forms, by "raising the index"

You could also raise an index on the metric tensor itself to transform it from a $$(0, 2)$$ tensor to a $$(1, 1)$$ tensor. The resulting linear map on vectors is boring: it's just the identity map. You can see this in components, because the process of raising an index uses the inverse metric, and you're left with the identity matrix.

By contrast, the strain tensor $$\epsilon_i^{\ \ j} = \partial_i u^j$$ is naturally a $$(1, 1)$$ tensor, and easy to interpret. If you contract a vector $$v^i$$ into it, then it simply returns the directional derivative of the displacement $$u^j$$ along the $$v^i$$ direction.