I begin my question with a multilinear question then I will consider two local smooth analogies:

Assume that $\alpha$ is a real valued symmetric $k$-tensor, that is a $k$-linear map $\alpha:\overbrace{\mathbb{R}^n\times\mathbb{R}^n\times\ldots\times \mathbb{R}^n}^{k\; \text{times}} \to \mathbb{R}$ which is unchanged under permutations.

Is there an (orthonormal) linear map $T$ on $\mathbb{R}^n$ such that $T^*(\alpha)$ is in the diagonal form $T^*(\alpha)=\sum_{i=1}^n \lambda_i \overbrace{dx_i\otimes dx_i\otimes \ldots \otimes dx_i}^{k \;\text{times}}$

We would like to consider two kind of smooth analogies of this multilinear question as follows:

Assume that $\omega$ is a smooth symmetric $k$ tensor on an open set of $\mathbb{R}^n$:

  1. What kind of obstructions would appear to represent $\omega$ in the form $\omega=\sum \overbrace{df_i \otimes df_i \otimes \ldots \otimes df_i}^{k \;\text{times}}$ for some smooth functions $f_1,f_2,\ldots,f_n$. When $k=2$ then this tensor is the pull back of the standard flat metric. Hence non vanishing of "Curvature" would be an obstruction for such a representation. What kind of tensors play the same role as curvature for $k>2$?

  2. What kind of obstructions would appear to have a change of coordinate $\phi$ such that $\phi^*{\omega}=\sum_{i=1}^n g_i \overbrace{dx_i\otimes dx_i\otimes \ldots \otimes dx_i}^{k\;\text{times}}$ for some smooth functions $g_1,g_2,\ldots, g_n$.


Any $k$-linear symmetric map $\alpha:(\mathbb R^n)^k \to \mathbb R$ is the polarization of a homogeneous polynomial $\bar \alpha: x\mapsto \alpha(x,x,x\dots,x)$ of degree $k$ on $\mathbb R^n$.

First question: You ask whether any homogeneous polynomial of degree $k$ is of the form $\lambda_1 x_1^k + \lambda_2 x_2^k +\dots+ \lambda_n x_n^k$ after a suitable orthogonal coordinate change on $\mathbb R^n$. Obviously, the answer is no.

Any decomposition into real irreducible factors would survive any coordinate transformation, for example. So $\{\bar\alpha = 0\}$ can be a union of cones, whose number and their degrees are invariants. You might want to assume that the polynomial $\bar\alpha$ is irreducible over the reals, for example. Then look at the real algebraic geometry classification of irreducible real codimension one varieties of degree $k$ in $\mathbb R\mathbb P^{n-1}$.


A point of view which might be more suited to your question: Determine first the orbit structure of the action of $GL(n)$ on the space $Pol^k[\mathbb R^n]$ of homogeneous polynomials of order $k$ on $\mathbb R^n$. If $k=1$ there is only 0 and $(\mathbb R^n)^*\setminus \{0\}$. If $k=2$ rank and signature are full invariants and each orbit consisting of non-degenerate inner products is open. If $k>2$, in general, you do not have open orbits by dimension reasons. So you have uncountably many types of symmetric tensor fields. Polynomial invariant theory does not help here since there are no polynomial $GL(n)$-invariants by Weyl's invariant tensor theorem. There are many $O(n)$-invariants, namely all polynomials in $g^{-1}(x_i,x_j)$, using the $O(n)$-invariant inner product $g$ on $\mathbb R^n$.

  • $\begingroup$ Can I ask you to explain your last sentence. Sorry if my question is elementary. $\endgroup$ – Ali Taghavi May 5 '18 at 16:25
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    $\begingroup$ @Ali Taghavi: For example, a product of linear factors remains a product of linear factors. Like $T^*(x_1^{k-p}x_2^p) = (T^*x_1)^{k-p}(T^*x_2)^p$. $\endgroup$ – Peter Michor May 5 '18 at 17:23

To your first question: no, even if we take out the orthonormal condition, by a dimension count argument (with one nontrivial exception).

Assume there is generally. The $\lambda_i$ can be removed by scaling the $x_i$ (possibly up to sign or degeneracy, neither of which affect dimension), so we can think about just the $x_i$. Each of the $\xi_i$ covering $x_i$ has $n$ degrees of freedom, so in total we have $n^2$ degrees of freedom to choose something of the form $\sum_{i=1}^n \pm \xi_i^{\otimes k}$.

But we are looking at elements of $\text{Sym}^k \mathbb{R}^n$, which has dimension $\frac{(n+k-1)!}{(n-1)!k!}$. This is too large (larger than $n^2$) unless $n=1$, or $k\leq 2$, or $n=2,k=3$.

From now on, I assume we are working over $\mathbb{C}$.

The first case is obvious: just scale the one $\xi$. The answer to both your questions for the smooth version is yes; just integrate the scaling factor. The second case is the usual spectral theorem on symmetric matrices. Its smooth version is the Riemann metric; for your second question, the obstruction is the Riemann tensor. For your third question, there is no obstruction.

It does turn out that this can be done (if we still discard the assumption of orthonormality) in the $n=2,k=3$ case (in fact, you can do this over any algebraically closed field of characteristic greater than $3$); in such a case, the choice of $\xi_i$ can be determined algebraically, up to reordering and scaling by $\zeta_3$.

As such, for the smooth version, assuming no degeneracy (in this case, that the tensor is nowhere the cube of a covector), by choosing one of the $6$ choices available locally at any point, we get $2$ everywhere linearly independent covector fields. An obstruction to the form in your second question for the tensor is equivalent to an obstruction to writing the covector fields as exterior derivatives of two functions. Obstructions for your second question are therefore the exterior derivatives of these two functions.

Obstructions for your third question are equivalent to obstructions to writing the covector fields as the product of a function and an exterior derivative of another function. This can be done for any nowhere zero covector fields on $\mathbb{R}^2$, so there is no obstruction.

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    $\begingroup$ If $k$ is even the sign of $\lambda_i$ cannot be skaled away. $\endgroup$ – Peter Michor May 5 '18 at 15:30
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    $\begingroup$ @PeterMichor you are correct; I was thinking of the argument over $\mathbb{C}$. I've now justified it a little more. $\endgroup$ – user44191 May 5 '18 at 15:38

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