1
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ 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? $\endgroup$ – fedja Jan 10 at 1:51
  • $\begingroup$ 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. $\endgroup$ – mr_e_man Jan 10 at 3:21
  • 1
    $\begingroup$ 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 :-) $\endgroup$ – fedja Jan 10 at 13:16
2
$\begingroup$

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 $$ |f(A)|\le \sum_{j,k,m,n}a_{jk}a_{km}a_{mn} a_{nj} \le \frac12\left(\sum_{j,k,m,n}a_{jk}^2a_{mn}^2 +\sum_{j,k,m,n}a_{km}^2 a_{nj}^2\right) = \left(\sum_{jk}a_{jk}^2\right)^2 = \|A\|^4. $$

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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). $\endgroup$ – mr_e_man Jan 13 at 17:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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