3
$\begingroup$

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

$\endgroup$
4
  • $\begingroup$ 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? $\endgroup$ Jan 27, 2019 at 15:28
  • $\begingroup$ 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 $1\leq j,k\leq n$. $\endgroup$
    – Student88
    Jan 27, 2019 at 16:25
  • 4
    $\begingroup$ No. Consider the $2 \times 2 \times 2$ cube with $t_{111} = t_{222} = 1$ and all other $t_{ijk} =0$. $\endgroup$ Jan 27, 2019 at 16:34
  • $\begingroup$ 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. $\endgroup$
    – Student88
    Jan 27, 2019 at 16:40

2 Answers 2

2
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ 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? $\endgroup$
    – Student88
    Jan 28, 2019 at 7:10
  • $\begingroup$ 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, $t_{ijk} t_{111}^2 = t_{ij1} t_{11k} t_{111} = t_{i11} t_{11k} t_{1j1}$. $\endgroup$ Jan 28, 2019 at 13:52
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.