Lemma 1 (Van der Corput inequality) To let be unit vectors in a Hilbert space . Then
Proof: The left side can be written as for some units complex numbers . We got from Cauchy-Schwarz
and the claim now follows from the triangle inequality.
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 be unit vectors in a Hilbert space so that for all and some < delta leq 1}" class="latex" />. Then we have for at least of couples .
Proof: From the lemma we have
The contribution of those With < delta ^ 2/2}" class="latex" /> is at most and all remaining summands are at most to give the claim.
One disadvantage of this inference is that it tells us nothing Which Couples are related to each other. Especially if the vector also correlates with a separate collection of unit vectors, the pairs for which The correlate may not have any intersection with the pairs in which correlate (except of course on the diagonal 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 be unit vectors in a Hilbert space for and so that for all , and some < delta_k leq 1}" class="latex" />. Then at least there is Couples so that . In particular (from Cauchy-Schwarz) we have for all .
Proof: Apply Corollary 2 to the unit vectors and , in the tensor force Hilbert space .
It is surprisingly difficult to get even a qualitative version of the above conclusion (namely, if correlates with all Then there are a lot of couples for which correlates with for all 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 for which one has correlation of , for a single 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
Where is the orthogonal projection to the complement of . This implies a gram matrix inequality
for each Where denotes the assertion that is positive semi-definite. We conclude from this with the Schur product set
and thus for a suitable selection of characters ,
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.