# Is the identification between symmetric tensors and homogeneous polynomials useful?

## The general question:

Given an $n$-dimensional vector space $V$ over a field $k$, there exists an identification $$\mathrm{Sym}^d(V) \sim k[x_1, \dots, x_n]_d$$ between the space of symmetric order $d$ tensors on $V$ and the space of order $d$ homogeneous polynomials in $n$ indeterminates over $k$.

I am wondering whether this identification is ever of much use in the study of tensors and tensor fields. In particular, I am interested in ways in which this identification might simplify problems in differential geometry, but uses in other fields would be interesting also.

This question is inspired by the following observation I made, which I would also like to confirm is valid:

Let $\omega \in \mathrm{Sym}^1(V)$ be some unknown. Suppose we have a map $$\phi \colon \mathrm{Sym}^1(V) \longrightarrow k$$ along with a few known elements $\eta, \theta, \nu_1, \nu_2 \in \mathrm{Sym}^1(V)$. Additionally, we have the following system of equations: \begin{align} \phi(\omega) \cdot \eta + \phi(\eta) \cdot \omega &= \nu_1 \\ \phi(\omega) \cdot \theta + \phi(\theta) \cdot \omega &= \nu_2 \end{align} Our goal is to solve for $\omega$. In light of the above, we can identify $\mathrm{Sym}^1(V)$ with $k[x_1, \dots, x_n]_1 \subset k[x_1, \dots, x_n]$, form the field of fractions $k(x_1, \dots, x_n)$ and solve the above system for $\omega$ using linear algebra. Doing so, we will arrive at an expression of the form: $$\left(\phi(\theta)\eta - \phi(\eta)\theta\right)\omega = \theta \cdot \nu_1 + \eta \cdot \nu_2$$ where all the elements are now considered to be in $k(x_1, \dots, x_n)$, so all the products make sense. Dividing, we can obtain an expression for $\omega$ expressed entirely in terms of objects we know. We can now evaluate this expression on the appropriate vectors to obtain an expression for $\omega$ in terms of a basis for $\mathrm{Sym}^1(V)$.

Note: I originally posted this question here on math.se. At the time I was unsure whether it would better to post it on mathoverflow. Given that it has remained unanswered over there for some time, I figured I would try posting it here. I am not as familiar with mathoverflow as some other stack exchange sites, so please feel free to edit my post in order to make things fit better with this site if necessary.

• Arguably, the only point to tensors is as homogeneous polynomials or multilinear functions. A tensor is difficult to interpret geometrically. Typically, it's only when you evaluate it on on the right number of tangent vectors (or cotangent vectors) that you get a number that you can interpret. The most obvious example of this is the Riemann curvature tensor. I don't know any way to explain the full tensor geometrically, but if you evaluate it properly using two tangent vectors (which span a 2-d plane), then you get sectional curvature which does mean something geometrically. – Deane Yang Sep 30 '14 at 16:53
• In Chern-Weil theory it's pretty useful. – user40276 Oct 1 '14 at 5:00
• @DeaneYang It sounds like you're saying that the tensor-polynomial identification is a useful tool within the context of differential geometry. However, does this identification provide a means to apply tools that may have been developed in other areas of mathematics to differential geometric problems? Said another way, might it be possible to rephrase problems from differential geometry as problems more like those encountered in a different branch of mathematics? – providence Oct 1 '14 at 5:01
• Sorry. I didn't read the question carefully. Your example is a question about tensors with respect to a single vector space, so it is a purely algebraic question. Differential geometry is effectively the study of tensors with respect to a smooth parameterized family of vector spaces. I don't see any need or use of differential geometry for a question like yours. – Deane Yang Oct 1 '14 at 17:51

I suggest an example showing that sometimes the object we need is actually the homogeneous polynomial in velocities but in order to work with it it is convenient to view it as a symmetric tensor since it allows us to use invariant ways to work with the tensors.

My example is the Killing tensor. In a simplest situation it is a symmetric $(0,2)$ tensor $K$ satisfying the equation $$symmetrisation \ of (\nabla K)=0.$$ If your prefer indices, the equation above looks $$K_{(ij,k)}=0.$$

The tensor plays improtant role in physics and differential geometry since it is corresponds to conservative quantaties of the geodesics flow: for a symmetric tensor $K$ the function $K(\dot \gamma, \dot \gamma)$ is constant along geodesics if and only if it is a Killing tensor. We see that the geometric condition that is equavalent to the property of tensor $K$ to be Killing is actually a condition about the symmetric polynomial $K(\xi, \xi)$.

Though the Killing equation above is equivalent to any other equation that is equivalent to the fact that $K(\dot\gamma, \dot\gamma)$ which is quadratic polynomial in $\dot \gamma$ is preserved along geodesics, writing it in the tensorial form above allows one to use the tensorial mashinery to work with this equation.

For example, one can use the tensorial mashinery to obtain conditions on the curvature that prevent a metric to have a Killing tensor. One can also use it in order to understand how many Killing tensors may exist on a manifold.

Also projective invariance of the Killing equations is better seenable on the level of Killing tensors and I did use it very extencively in my research.

Everything what I said is also true for higher degree $d$ of the polynomial, i.e., for higher valency of the tensor.