# Is this function on antisymmetric 3-tensors (over $\ell^2$) bounded? $\sum_{i,j,k,l,m,n}A_{ijk}A_{imn}A_{ljn}A_{lmk}$

Let $$A$$ be an antisymmetric 3-tensor over a sequence space (for simplicity, assume finite sequences):

$$A_{ijk}=-A_{jik}=-A_{ikj}\in\mathbb R,\quad(i,j,k)\in\mathbb N^3$$

$$\lVert A\rVert^2=\sum_{i,j,k}|A_{ijk}|^2<\infty$$

and consider the function

$$f(A)=\sum_{i,j,k,l,m,n}A_{ijk}A_{imn}A_{ljn}A_{lmk}$$

$$=\sum_{j,k,m,n}\left(\sum_iA_{ijk}A_{imn}\right)\left(\sum_lA_{ljn}A_{lmk}\right).$$

Is there a constant $$C>0$$ such that $$|f(A)|\leq C\lVert A\rVert^4$$ for all $$A$$?

I tried using the Cauchy-Schwarz inequality in different ways.

• It looks like for any two symmetric real matrices $A,B$, we have $Tr[ABAB]\le Tr[A^2]Tr[B^2]$ and it seems like this has been proved somewhere on MO or MSE already. This would settle the question with constant $C=1$ but I have hard time finding the link. Can anybody help? – fedja Jan 10 at 1:51
• With the Hilbert-Schmidt norm and inner product, that is $$\text{tr}(ABAB)=\text{tr}(A^TB^TAB)=\text{tr}((BA)^T(AB))=\langle BA,AB\rangle\leq\lVert BA\rVert\lVert AB\rVert$$ $$\leq\lVert B\rVert\lVert A\rVert\lVert A\rVert\lVert B\rVert=\lVert A\rVert^2\lVert B\rVert^2=\text{tr}(A^TA)\text{tr}(B^TB)=\text{tr}(A^2)\text{tr}(B^2).$$ But I don't see how this relates to my question. – mr_e_man Jan 10 at 3:21
• You can just fix $k$ and $n$ and (replacing everything with absolute values and using symmetry) apply the inequality I mentioned to other indices, after which the summation over $k$ and $n$ will become easy. However daw gave you an excellent answer already, so I'll not go into details :-) – fedja Jan 10 at 13:16

First, $$\left|\sum_i A_{ijk}A_{imn}\right| \le \sum_i |A_{ijk}|\cdot |A_{imn}| \le \left(\sum_i |A_{ijk}|^2\right)^{1/2} \left(\sum_i |A_{imn}|^2\right)^{1/2} .$$ Define $$a_{jk}^2:= \sum_i |A_{ijk}|^2.$$ Then $$|f(A)|\le \sum_{j,k,m,n}a_{jk}a_{km}a_{mn} a_{nj} \le \frac12\left(\sum_{j,k,m,n}a_{jk}^2a_{mn}^2 +\sum_{j,k,m,n}a_{km}^2 a_{nj}^2\right) = \left(\sum_{jk}a_{jk}^2\right)^2 = \|A\|^4.$$
• Thanks @daw. This simple application of the AM-GM inequality was the last thing I needed, in order to prove that multiplication $AB$ in the Clifford algebra is continuous/bounded when the grades of $A$ and $B$ are bounded. This particular question was for the quadvector (grade 4) part of $A^2$ when $A$ is a trivector (grade 3). – mr_e_man Jan 13 at 17:53