Rank of order-3 tensor with all slices being rank-1

Question

If some tensor $T=(t_{ijk})$ has that all of its (2 dimensional) slices (along all 3 axes) are of rank-1, does it follow that the tensor is also rank-1? That is, can be written as $$ t_{ijk}=a_i b_j c_k$$ ?

Answer 1

Let $t$ be a nonzero tensor. Then some $t_{ijk}$ are nonzero, without loss of generality let $t_{111}\neq 0$. Rescaling our tensor, we may assume that $t_{111}=1$. Put $a_i = t_{i11}$, $b_j = t_{1j1}$ and $c_k = t_{11k}$. Then $t$ is rank $1$ if and only if $t_{ijk} = a_i b_j c_k$.

If all $t_{11k}$ are nonzero, then this follows from the condition on $2$-dimensional slices: $$(t_{ijk} t_{11k}) t_{111}^2 = (t_{i1k} t_{1jk}) t_{111}^2 = (t_{i1k} t_{111}) (t_{1jk} t_{111}) = (t_{i11} t_{11k}) (t_{1j1} t_{11k}) = (t_{i11} t_{1j1} t_{11k}) t_{11k} $$ so $$t_{ijk} t_{11k} = a_i b_j c_k t_{11k}.$$ One could obviously use conditions that various $t_{1j1}$ or $t_{i11}$ are nonzero instead.

Without any nonvanishing assumptions, the easiest thing I can see to do is to impose a condition on slanted $2 \times 2$ blocks: $t_{i_1 j_1 k_1} t_{i_2 j_2 k_2} = t_{i_1 j_1 k_2} t_{i_2 j_2 k_1}$ and likewise in switching the $i$-indices and the $j$-indices.

Answer 2

The following comments may be useful to you, though you may not regard them as a complete answer.

To set the stage, first consider the case of an order 2 tensor $T\in V_1\otimes V_2$, where $V_i$ are vector spaces of dimension $n_i\ge 2$. if $(e^1,\ldots, e^{n_1})$ is a basis of $V_1$ and $(f^1,\ldots, f^{n_2})$ is a basis of $V_2$, then we can write $$ T = t_{ij}\,e^i{\otimes}f^j, $$ and the sufficient condition that $T$ be rank $1$ is that $t_{i_1j_1}t_{i_2j_2}- t_{i_1j_2}t_{i_2j_1}=0$, i.e., that the tensor $$ T^{[2]} = (t_{i_1j_1}t_{i_2j_2}- t_{i_1j_2}t_{i_2j_1})\, e^{i_1}{\wedge}e^{i_2}\otimes f^{j_1}{\wedge}f^{j_2} $$ should vanish. Note that this is ${n_1\choose2}{n_2\choose2}$ distinct quadratic equations, which seems, at first glance, to be far more equations than necessary to cut out the rank-at-most-1 locus $R_1\subset V_1\otimes V_2$, which is a cone of dimension $n_1+n_2-1$ in a vector space of dimension $n_1n_2$ (and hence, one might expect to be able to do it with just $(n_1{-}1)(n_2{-}1)$ equations). Of course, if you know a priori that some $t_{ij}\not=0$, then the $(n_1{-}1)(n_2{-}1)$ equations $$ t_{i'j}t_{ij'} - t_{ij}t_{i'j'} = 0,\quad\text{where}\quad i'\not=i,\ \ j'\not=j\, $$ do suffice to describe $R_1$ on the open set where $t_{ij}\not=0$. However, in the hyperplane $H_{ij}\subset V_1\otimes V_2$ defined by $t_{ij}=0$, the above equations do not describe $R_1\cap H_{ij}$.

Now, you might want to find some subset of the components of $T^{[2]}$ that will do the job, but there is no such subset that is invariant under change of basis in the two vector spaces, i.e., under $\mathrm{GL}(V_1)\times\mathrm{GL}(V_2)$. The reason is that the quadratic polynomials on $V_1\otimes V_2$ are given by $S^2\bigl((V_1\otimes V_2)^*\bigr)=S^2(V_1^*\otimes V_2^*)$, and it is well-known that, as a $\mathrm{GL}(V_1)\times\mathrm{GL}(V_2)$-module, we have a decomposition into two irreducible subspaces: $$ S^2(V_1^*\otimes V_2^*) = S^2(V_1^*){\otimes}S^2(V_2^*)\oplus \Lambda^2(V_1^*){\otimes}\Lambda^2(V_2^*). $$ It is the polynomials in the second irreducible subspace, $\Lambda^2(V_1^*){\otimes}\Lambda^2(V_2^*)$, that are the coefficients of $T^{[2]}$, and, obviously, there are ${n_1\choose2}{n_2\choose2}$ of them, and they are linearly independent.

Now, in the case of tensors of order $3$, we have three vector spaces to deal with, and the corresponding module decomposition has four irreducible components: $$ S^2(V_1^*\otimes V_2^*\otimes V_3^*) = S^2(V_1^*){\otimes}S^2(V_2^*){\otimes}S^2(V_3^*)\oplus W_1\oplus W_2 \oplus W_3 $$ where, for $i, j, k$ distinct, we have $$ W_i = S^2(V_i^*)\otimes \Lambda^2(V_j^*) \otimes \Lambda^2(V_k^*). $$ Now, it's easy to check that the quadratic functions on $V_1\otimes V_2\otimes V_3$ that correspond to the elements of $W_i$ (for $i = 1, 2, 3$) all vanish on $R_1\subset V_1\otimes V_2\otimes V_3$. Moreover, it's easy to find 3-tensors of rank greater than $1$ such that all the elements of $W_i$ and $W_j$ vanish on them for any distinct pair $(i,j)$. Hence, the only subspace of $S^2\bigl((V_1{\otimes}V_2{\otimes}V_3)^*\bigr)$ that is invariant under $\mathrm{GL}(V_1)\times\mathrm{GL}(V_2)\times\mathrm{GL}(V_2)$ that could possibly define $R_1\subset V_1\otimes V_2\otimes V_3$ is the subspace $I = W_1{\oplus}W_2{\oplus}W_3$. Meanwhile, it turns out that this subspace does indeed define $R_1$, in the sense that a 3-tensor $T$ has rank at most 1 if and only if all of the elements of $I$ vanish on it.

Of course, the dimension of $I$ is very large (roughly $\tfrac38(n_1n_2n_3)^2$), far larger than the codimension of $R_1$ in $V_1\otimes V_2\otimes V_3$. Thus, using $I$ to test 3-tensors for membership in $R_1$ is highly inefficient. As David has already shown, you do not need to check the vanishing of all the quadratics in $I$ on a given $T = t_{ijk}\, e^i{\otimes}f^j{\otimes}g^k$ if you know that certain particular sets of components of $T$ are nonzero.

However, you will have to check at least $n_1n_2n_3-(n_1+n_2+n_3-2)$ independent conditions, since that is the codimension of $R_1$ in $V_1\otimes V_2\otimes V_3$. You can judge how good a given criterion/algorithm is by checking how far you are from this obvious lower bound. What's certain is that there is no `natural' (i.e., independent of basis change) proper subspace of $I$ that will work, so you will have to give up this 'naturality' to get a more efficient algorithm.