In this previous blog post I noticed the following simple application of Cauchy Black:

Lemma 1 (Van der Corput inequality) To let ${v, u_1, dots, u_n}$ be unit vectors in a Hilbert space ${H}$. Then $displaystyle ( sum_ {i = 1} ^ n | langle v, u_i rangle_H |) ^ 2 leq sum_ {1 leq i, j leq n} | langle u_i, u_j rangle_H |.$

Proof: The left side can be written as ${| langle v, sum_ {i = 1} ^ n epsilon_i u_i rangle_H | ^ 2}$ for some units complex numbers ${ epsilon_i}$. We got from Cauchy-Schwarz $displaystyle | langle v, sum_ {i = 1} ^ n epsilon_i u_i rangle_H | ^ 2 leq langle sum_ {i = 1} ^ n epsilon_i u_i, sum_ {j = 1} ^ n epsilon_j u_j rangle_H$

and the claim now follows from the triangle inequality. $Box$

Hence, the correlation becomes transitive in the statistical sense (although it is not transitive in the absolute sense):

Corollary 2 (statistical transitivity of the correlation) To let ${v, u_1, dots, u_n}$ be unit vectors in a Hilbert space ${H}$ so that ${| langle v, u_i rangle_H | geq delta}$ for all ${i = 1, dots, n}$ and some < delta leq 1}" class="latex" />. Then we have ${| langle u_i, u_j rangle_H | geq delta ^ 2/2}$ for at least ${ delta ^ 2 n ^ 2/2}$ of couples ${(i, j) in {1, dots, n } ^ 2}$.

Proof: From the lemma we have $displaystyle sum_ {1 leq i, j leq n} | langle u_i, u_j rangle_H | geq delta ^ 2 n ^ 2.$

The contribution of those ${i, j}$ With < delta ^ 2/2}" class="latex" /> is at most ${ delta ^ 2 n ^ 2/2}$and all remaining summands are at most ${1}$to give the claim. $Box$

One disadvantage of this inference is that it tells us nothing Which Couples ${u_i, u_j}$ are related to each other. Especially if the vector ${v}$ also correlates with a separate collection ${w_1, dots, w_n}$ of unit vectors, the pairs ${(i, j)}$ for which ${u_i, u_j}$ The correlate may not have any intersection with the pairs in which ${w_i, w_j}$ correlate (except of course on the diagonal ${i = j}$ where they need to correlate).

While working on an ongoing research project, I recently discovered that there is a very simple way to get around the latter problem by taking advantage of the tensor power trick:

Corollary 3 (Simultaneous statistical transitivity of the correlation) To let ${v, u ^ k_i}$ be unit vectors in a Hilbert space for ${i = 1, dots, n}$ and ${k = 1, dots, K}$ so that ${| langle v, u ^ k_i rangle_H | geq delta_k}$ for all ${i = 1, dots, n}$, ${k = 1, dots, K}$ and some < delta_k leq 1}" class="latex" />. Then at least there is ${( delta_1 dots delta_K) ^ 2 n ^ 2/2}$ Couples ${(i, j) in {1, dots, n } ^ 2}$ so that ${ prod_ {k = 1} ^ K | langle u ^ k_i, u ^ k_j rangle_H | geq ( delta_1 dots delta_K) ^ 2/2}$. In particular (from Cauchy-Schwarz) we have ${| langle u ^ k_i, u ^ k_j rangle_H | geq ( delta_1 dots delta_K) ^ 2/2}$ for all ${k}$.

Proof: Apply Corollary 2 to the unit vectors ${v ^ { otimes K}}$ and ${u ^ 1_i otimes dots otimes u ^ K_i}$, ${i = 1, dots, n}$ in the tensor force Hilbert space ${H ^ { otimes K}}$. $Box$

It is surprisingly difficult to get even a qualitative version of the above conclusion (namely, if ${v}$ correlates with all ${u ^ k_i}$Then there are a lot of couples ${(i, j)}$ for which ${u ^ k_i}$ correlates with ${u ^ k_j}$ for all ${k}$ simultaneously) without any version of the tensor power trick. For example, even the powerful Szemerédi regularity lemma, if it is applied to the set of pairs ${i, j}$ for which one has correlation of ${u ^ k_i}$, ${u ^ k_j}$ for a single ${i, j}$does not seem to be sufficient. However, there is a reformulation of the argument using the Schur product set as a replacement (or really a disguised version) of the tensor power trick. To simplify the notation, let’s just work with real Hilbert spaces to illustrate the argument. We start with identity $displaystyle langle u ^ k_i, u ^ k_j rangle_H = langle v, u ^ k_i rangle_H langle v, u ^ k_j rangle_H + langle pi (u ^ k_i), pi (u ^ k_j) rangle_H$

Where ${Pi}$ is the orthogonal projection to the complement of ${v}$. This implies a gram matrix inequality $displaystyle ( langle u ^ k_i, u ^ k_j rangle_H) _ {1 leq i, j leq n} succ ( langle v, u ^ k_i rangle_H langle v, u ^ k_j rangle_H) _ {1 leq i, j leq n} succ 0$

for each ${k}$ Where ${A succ B}$ denotes the assertion that ${FROM}$ is positive semi-definite. We conclude from this with the Schur product set $displaystyle ( prod_ {k = 1} ^ K langle u ^ k_i, u ^ k_j rangle_H) _ {1 leq i, j leq n} succ ( prod_ {k = 1} ^ K langle v, u ^ k_i rangle_H langle v, u ^ k_j rangle_H) _ {1 leq i, j leq n}$

and thus for a suitable selection of characters ${ epsilon_1, dots, epsilon_n}$, $displaystyle sum_ {1 leq i, j leq n} epsilon_i epsilon_j prod_ {k = 1} ^ K langle u ^ k_i, u ^ k_j rangle_H geq delta_1 ^ 2 dots delta_K ^ 2 n ^ 2.$

One now argues as in the proof of Corollary 2.

A separate application of tensor forces to reinforce correlations was also mentioned in this previous blog post which features a cheap version of the Kabatjanskii-Levenstein bond, but this does not seem to be directly related to this current application.