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

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 $$t_{ijk}=a_i b_j c_k$$ ?

• When you say that 2-dimensional slices are of rank 1, do you mean that its image under $\mathrm{id} \otimes \mathrm{id} \otimes \xi$ is of rank 1 for all linear functionals $\xi : V \to k$, or for only linear functionals $\xi$ that are projection to the axes? Jan 27, 2019 at 15:28
• The later I think. That is, all the matrices that are received by “freezing” one index. For example the matrix $A=(a_{ijk})$, for $i=3$,and $1\leq j,k\leq n$. Jan 27, 2019 at 16:25
• No. Consider the $2 \times 2 \times 2$ cube with $t_{111} = t_{222} = 1$ and all other $t_{ijk} =0$. Jan 27, 2019 at 16:34
• Thank you @DavidESpeyer. Is there an equivalent condition for that a tensor is rank-1? For example, in the matrix case, I know that if all $2\times 2$ minors vanish (equal to 0), then the matrix is rank-1. I'm looking for a similar condition in the 3-order tensor case. Jan 27, 2019 at 16:40

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.

• Thanks, but I didn't quite understand how did you derive this condition. I do know that for any 6 indices $(i,j,k,p,g,l)$ we got $t_{ijk}t_{pgl}=t_{ijl}t_{pgk}$. But why does this imply that $T$ is rank-1? Jan 28, 2019 at 7:10
• To repeat: One of the $t_{ijk}$ is nonzero, without loss of generality $t_{111}$. Rescaling, $t_{111}=1$. Put $a_i = t_{i11}$, $b_j = t_{1j1}$ and $c_k = t_{11k}$. Then we checked that $t_{ijk} = a_i b_j c_k$. In terms of the $2 \times 2$ conditions, $t_{ijk} t_{111}^2 = t_{ij1} t_{11k} t_{111} = t_{i11} t_{11k} t_{1j1}$. Jan 28, 2019 at 13:52

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.