# Operator norm of sum of tensor products

Let $$A,B,C,D$$ be bounded operators on a Hilbert space $$\mathcal{H}$$. I know that $$\|AB\| \leq \|A\otimes B\|$$ where $$\otimes$$ is the tensor product and $$\|\cdot\|$$ is the operator norm. I wonder if this extends to linear combinations. That is, is the following generalization true: $$\|AB + CD\| \leq \|A\otimes B + C\otimes D\| ?$$

My intuition is that it is harder'' for $$A\otimes B$$ to cancel $$C\otimes D$$ than it is for $$AB$$ to cancel $$CD$$, so I think it might be true.

• Pretty sure this is false. I know for certain this is false if instead of two pairs, we consider arbitrary finitely many pairs, though I can’t really come up with an explicit counterexample at the moment. Commented Aug 11 at 11:49
• Where can I find a reference for the counterexample with finitely many pairs? That already would be interesting to me. Commented Aug 11 at 12:49
• You already have Severin’s answer. What I had in mind was when $A, C$ commute with $B, D$. Then the question for finitely many pairs becomes about nuclearity of the algebras generated by $A, C$ and by $B, D$. The result then follows by the existence of non-nuclear $C^\ast$-algebras. Commented Aug 11 at 13:05
• @DavidGao Would you mind spelling this out as an answer (if it is not too tedious)? That sounds quite interesting. Commented Aug 11 at 13:17
• @SeverinSchraven Done. Commented Aug 11 at 13:42

Let's assume that the Hilbert space has at least dimension $$2$$ (in dimension $$1$$ the inequality is obviously true) and pick three orthonormal vectors $$e_1, e_2$$. Then consider the rank $$1$$ projections $$A= \vert e_1 \rangle \langle e_1 \vert, B= \vert e_1 \rangle \langle e_1 \vert, C= \vert e_1 \rangle \langle e_2 \vert, D = \vert e_2 \rangle \langle e_1 \vert.$$ Then we get $$AB+CD=2 \vert e_1 \rangle \langle e_1 \vert.$$ Clearly this has operator norm $$2$$. Extend our family to a Hilbert basis $$(e_j)_{j\in \Lambda}$$ of $$H$$. Then we compute $$(A\otimes B+C\otimes D)(\sum_{i,j\in \Lambda} c_{i,j} \, e_i \otimes e_j) = c_{1,1} \, e_1 \otimes e_1+ c_{2,1} \, e_1 \otimes e_2.$$ By orthogonality we get $$\Vert c_{1,1} \, e_1 \otimes e_1+ c_{2,1} \, e_1 \otimes e_2 \Vert = \sqrt{\vert c_{1,1}\vert^2+\vert c_{2,1}\vert^2} \leq \left( \sum_{i,j\in \Lambda} \vert c_{i,j}\vert^2\right)^{1/2} = \Vert \sum_{i,j\in \Lambda} c_{i,j} \, e_i \otimes e_j \Vert.$$ Thus, $$\Vert A\otimes B + C \otimes D \Vert =1$$ and hence $$\Vert AB+CD \Vert =2 > 1 = \Vert A\otimes B + C \otimes D \Vert.$$

• I like this simple example. I suppose my intuition was the $A\otimes B$ would have a harder time cancelling (destructively) with $C\otimes D$, but what this shows is that the converse is also true -- $A\otimes B$ has a harder time "constructively interfering" with $C\otimes D$. Commented Aug 11 at 13:53
• @felipeh FWIW, I think your intuition is nearly correct. While the norm inequality does not hold, it is still the case that $A \otimes B + C \otimes D = 0$ would imply $AB + CD = 0$. Commented Aug 11 at 13:57

As requested by @SeverinSchraven, here is an elaboration of my comments. What I had in mind was something like the following:

Consider two $$C^\ast$$-algebras $$\mathcal{A}$$, $$\mathcal{B}$$. In $$C^\ast$$-algebra theory, there are, in general, multiple norms that can be equipped on the tensor product $$\mathcal{A} \otimes \mathcal{B}$$ to make it, after completion, into a $$C^\ast$$-algebra. Two canonical choices are the smallest, minimal (also called spatial) tensor norm $$\otimes_\min$$; and the largest, maximal tensor norm $$\otimes_\max$$. A $$C^\ast$$-algebra $$\mathcal{A}$$ is called nuclear if for all $$C^\ast$$-algebra $$\mathcal{C}$$, the minimal and maximal tensor norms on $$\mathcal{A} \otimes \mathcal{C}$$ coincide. It is a classic result that there are non-nuclear $$C^\ast$$-algebras, for example the reduced group $$C^\ast$$-algebra $$C^\ast_r(G)$$ (or full group $$C^\ast$$-algebra $$C^\ast(G)$$) of a discrete, non-amenable group $$G$$ (for example any non-abelian free group). Also, $$\mathbb{B}(l^2)$$ is non-nuclear. Regardless, the point is there exists some $$C^\ast$$-algebras $$\mathcal{A}$$ and $$\mathcal{B}$$ s.t. $$\mathcal{A} \otimes_\min \mathcal{B} \neq \mathcal{A} \otimes_\max \mathcal{B}$$. However, basically by definition, there always exists a contractive $$\ast$$-homomorphism $$\mathcal{A} \otimes_\max \mathcal{B} \to \mathcal{A} \otimes_\min \mathcal{B}$$ which is the identity map on the algebraic tensor product $$\mathcal{A} \otimes \mathcal{B}$$. $$\mathcal{A} \otimes_\min \mathcal{B} \neq \mathcal{A} \otimes_\max \mathcal{B}$$ then means there are some $$A_1, \cdots, A_n \in \mathcal{A}$$ and $$B_1, \cdots, B_n \in \mathcal{B}$$ s.t. $$\|\sum_{i=1}^n A_i \otimes B_i\|_{\mathcal{A} \otimes_\min \mathcal{B}} < \|\sum_{i=1}^n A_i \otimes B_i\|_{\mathcal{A} \otimes_\max \mathcal{B}}$$.

Let $$H$$ be a Hilbert space on which $$\mathcal{A} \otimes_\max \mathcal{B}$$ acts faithfully. Both $$\mathcal{A}$$ and $$\mathcal{B}$$ canonically (and isometrically) embeds into $$\mathcal{A} \otimes_\max \mathcal{B}$$, so we can regard $$A_i$$, $$B_i$$ as operators on $$H$$. Then,

$$\|\sum_{i=1}^n A_i \otimes B_i\|_{\mathcal{A} \otimes_\max \mathcal{B}} = \|\sum_{i=1}^n A_iB_i\|_{\mathbb{B}(H)}$$

On the other hand, again basically by definition of the minimal tensor norm,

$$\|\sum_{i=1}^n A_i \otimes B_i\|_{\mathcal{A} \otimes_\min \mathcal{B}} = \|\sum_{i=1}^n A_i \otimes B_i\|_{\mathbb{B}(H \otimes H)}$$

So you have $$\|\sum_{i=1}^n A_i \otimes B_i\|_{\mathbb{B}(H \otimes H)} < \|\sum_{i=1}^n A_iB_i\|_{\mathbb{B}(H)}$$.

• Thanks a lot! I feel it is helpful to have different points of view. Commented Aug 11 at 13:48
• Thank you for this answer! I am confused about your equation for $\otimes_{\text{max}}$. If $A$ and $B$ are nonzero and $AB=0$, doesn't this seems to imply that $\|A\otimes B\|_{\mathcal{A}\otimes_{\text{max}}\mathcal{B}} = 0$? Or is the point that there are no such $A$ and $B$ in $C^*$ algebras? I am very much not an operator theorist so I'm aware this may be a very silly question. Commented Aug 11 at 13:50
• @felipeh Admittedly, the notation here is slightly confusing. The point is $A$ and $B$ are originally in two different algebras, $\mathcal{A}$, $\mathcal{B}$, and the multiplication wouldn’t have made sense. But I’m representing $\mathcal{A}\otimes_\max\mathcal{B}$ on a Hilbert space $H$ and regard $\mathcal{A}$ and $\mathcal{B}$ as subalgebras of $\mathcal{A}\otimes_\max\mathcal{B}$. For simplicity, assume these algebras are unital, then when I wrote $A$, I really meant $A \otimes 1$ in $\mathcal{A}\otimes_\max\mathcal{B}$. Similarly, $B$ really meant $1 \otimes B$ in… Commented Aug 11 at 13:54
• @felipeh … $\mathcal{A}\otimes_\max\mathcal{B}$. Then $AB$ is really $A \otimes B$ in $\mathcal{A}\otimes_\max\mathcal{B}$. Of course, this cannot be zero unless $A$ or $B$ is zero. Commented Aug 11 at 13:55