Sylvester's Criterion for tensors We know from linear algebra that if $A = (a_{ij})_{i,j=1}^n$ is a symmetric matrix, then $A$ is positive-definite if and only if all principal minors of $A$ are positive. 
If I say that a symmetric (covariant) $k$-tensor $(T_{i_1\cdots i_k})_{i_1,\ldots, i_k=1}^n$ is positive-definite if for any non-zero vector $v = (v^1,\ldots, v^n)$, we have that $$\sum T_{i_1\cdots i_k}v^{i_1}\cdots v^{i_k} > 0,$$is there a Sylvester-like criterion for that? I imagine so, but it should be ugly. I know no references, though.
 A: By polarization, a symmetric $k$-tensor is the same as a homogeneous polynomial of degree $k$ on $\mathbb{R}^n$. If $T = T_{i_1, \dots, i_k} dx^{i_1} \otimes \cdots \otimes dx^{i_k}$ then the corresponding homogeneous polynomial is given by
$$ p(v) = T(v, \cdots, v) $$
or, more explicitly,
$$ p(v^1, \cdots, v^n) "=" p(v^i e_i) = T(v^{j_1} e_{j_1}, \dots, v^{j_k} e_{j_k}) = T_{i_1 \cdots i_k} dx^{i_1}(v^{j_1} e_{j_1}) \cdots dx^{i_k}(v^{j_k}e_{j_k}) = T_{i_1 \cdots i_k} v^{i_1} \cdots v^{i_k}.$$
Hence, your question is equivalent to the question of whether the polynomial $p$ is positive definite in the sense that if $v \neq 0$ then $p(v) > 0$. 
When $k$ is odd $p$ clearly cannot be positive definite: if $p(v) > 0$ for some $v$ then 
$$p(-v) = (-1)^k p(v) = -p(v) < 0.$$
When $k$ is even, as far as I know there aren't any effective criteria except maybe in special cases. See here, here and here (theorem 3.3) for some partial results and further references. 
By the way, this is one of the reasons one usually doesn't bother discussing higher order derivative tests for minimum/maximum of a function except when $n = 1$. If you have a multivariable function $f \colon \mathbb{R}^n \rightarrow \mathbb{R}$ such that 
$$(\nabla f)(x_0) = 0, \cdots, (\nabla^{k-1} f)(x_0) = 0$$
and if $(\nabla^k f)(x_0)$ (identified with a homogenous polynomial of degree $k$, the degree $k$ part of the Taylor expansion of $f$) is positive definite then $x_0$ will be a minimum point. But when $k > 2$, it seems we just don't have an "explicit" criterion to determine if this is the case or not.
