# Is the interior of the tensor product of two convex cones equal to the tensor product of their respective interiors?

I am sorry that the following question is elementary. I have not received an answer from my post at Math Stack Exchange.

In the following question, all cones are convex and contain the origin. Let $$C \subset \mathbb{R}^{m}$$ be a cone consisting of $$m$$-dimensional column vectors, and $$D \subset \mathbb{R}^{n}$$ another cone of $$n$$-dimensional column vectors.

Given any subsets $$A \subset \mathbb{R}^{m}$$ and $$B\subset \mathbb{R}^{n}$$, define $$A \otimes B \subset \mathbb{R}^{m \times n}$$ as the set of $$m \times n$$ matrices of the form $$x_1 {y_1}^T + \dotsb + x_q {y_q}^T$$, where $$x_1, \dotsc, x_q \in A$$ and $$y_1, \dotsc, y_q \in B$$, $$q\geq 1$$ is an arbitrary positive integer, and $$^T$$ denotes the transpose.

Is it true that $$(C \otimes D)^\circ = C^\circ \otimes D^\circ,$$ where $$^\circ$$ denotes the interior relative to the respective euclidean topologies?

Here's an example where the said equality holds. Take $$C = {\mathbb{R}_{\ge 0}}^m$$ to be the cone of all $$m$$-dimensional column vectors whose every entry is a nonnegative real number. Similarly, take $$D = {\mathbb{R}_{\ge 0}}^n$$. Then $$C \otimes D = {\mathbb{R}_{\ge 0}}^{m \times n}$$ is the cone of $$m \times n$$ matrices whose every entry is nonnegative. Indeed, it suffices to show that $$\mathbf{E}_{ij}$$, the matrix whose $$(i, j)$$th entry is $$1$$ and has $$0$$s everywhere else, lies in $$C \otimes D$$, because $$C \otimes D$$ is closed under taking nonnegative linear combinations. But $$\mathbf{E}_{ij} = \vec{e_i}\vec{e_j}^T$$, where $$\vec{e_i} \in C$$ and $$\vec{e_j} \in D$$. Therefore $$(C \otimes D)^\circ = {\mathbb{R}_{> 0}}^{m \times n}$$ is the set of all $$m \times n$$ matrices whose every entry is strictly positive.

Noting that $$C^\circ = {\mathbb{R}_{> 0}}^m$$ consists of all $$m$$-dimensional vectors whose entries are all strictly positive real numbers, and similarly $$D^\circ = {\mathbb{R}_{> 0}}^n$$, we get $$C^\circ \otimes D^\circ = {\mathbb{R}_{> 0}}^{m \times n}$$ also.

• If $x_1\in C$ and $y_1\in D$, then $x_1$ is $m\times1$ and $y_1^T$ is $n\times1$. What can then $x_1y_1^T$ mean? Apr 17, 2023 at 20:20
• Iosif, thanks for pointing out my error. The transpose should not be there. Apr 18, 2023 at 4:48

Yes, $$(C \otimes D)^\circ = C^\circ \otimes D^\circ$$ is correct.
Let us start by proving the inclusion $$C^\circ \otimes D^\circ \subseteq (C \otimes D)^\circ$$. To this end, it is enough to show that $$C^\circ \otimes D^\circ$$ is open. For a given point $$z = x_1 {y_1}^T + \dotsb + x_q {y_q}^T \in C^\circ \otimes D^\circ$$, we can use convexity and openness to replace $$x_1$$ by a convex combination of $$n$$ linearly independent points in $$C$$, and similarly for $$y_1$$. In this way, we can assume without loss of generality that the $$x_i$$'s span $$\mathbb{R}^m$$, and similarly the $$y_i$$'s span $$\mathbb{R}^n$$. Therefore by nudging each one of these points a little (without leaving $$C^\circ$$ or $$D^\circ$$), we can move in any direction in $$C^\circ \otimes D^\circ$$, and it follows that the latter set is open.
Furthermore, the inclusion $$C^\circ \otimes D^\circ \subseteq (C \otimes D)^\circ$$ is dense, because $$C^\circ \otimes D^\circ$$ is dense even in $$C \otimes D$$ by the obvious termwise approximation argument.
To now arrive at the claimed equality, recall that every convex open set is the interior of its closure. This applies in particular to both $$C^\circ \otimes D^\circ$$ and $$(C \otimes D)^\circ$$. This gives $$C^\circ \otimes D^\circ = \overline{C^\circ \otimes D^\circ}^\circ = \overline{(C \otimes D)^\circ}^\circ = (C \otimes D)^\circ,$$ where the second step is the density from the previous paragraph.
• For the first paragraph, to prove that $C^\circ \otimes D^\circ$ is open, do we actually need the stronger assumption that $x_i \otimes y_j = x_i {y_j}^T$ span $\mathbb{R}^n \otimes \mathbb{R}^m \cong \mathbb{R}^{m \times n}$? Apr 27, 2023 at 6:51
• I was trying to write out the argument explicitly. Say the nudging is in the direction $x \otimes y$, where $x = \sum_i \lambda_i x_i$ and $y = \sum_j \mu_j y_j$ for scalars $\lambda_i, \mu_j$. Then $x \otimes y = \sum_{i, j} \lambda_i \mu_j x_i \otimes y_j$. So what do I do with the cross-terms $x_i \otimes y_j$ where $i \neq j$? Apr 27, 2023 at 6:53
• Ah sorry, I think my formulation of that didn't fully capture what I actually meant. What I meant is that we can assume without loss of generality that $z = w + \sum_{i,j} \beta_{i,j} x_i \otimes y_j$, where the $x_i$ and the $y_j$ are bases, $\beta_{i,j} > 0$ for all $i$ and $j$, and $w$ contains all the remaining terms. Is this clearer? Then you can move in all direction simply by perturbing the coefficients $\beta_{i,j}$. Apr 27, 2023 at 12:21