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Consider the set $ X_1^n=\{1,2,...,2^n\} $. Then define $ X_2^n $ to be the set of two element subsets of $ X_1^n $. I will construct $ X_i $ by induction on $ i $. $ X_i^n $ is the set of two element subsets of $ X_{i-1}^n $ so that those two elements are disjoint, i.e. $ X_i^n=\{\{x,y\}| x,y\in S_{i-1}, x\cap y=\phi\} $. Here by intersection I mean that, we have a map $ \phi_i $ from $ X_i^n $ to subsets of $ X_1^n $ which is defined by induction as follows:- $ \phi_{i+1}(\{x,y\})=\phi_i(x)\cup \phi_i(y) $. So by $ x\cap y $ I meant $ \phi_{i-1}(x)\cap \phi_{i-1}(y) $. For example, an element of $ X_3^n $ will look something like this:- $ \{\{1,2\},\{3,4\}\} $. An element of $ X_4^n $ will look like $ \{\{\{1,3\},\{2,5\}\},\{\{4,7\},\{6,8\}\}\} $. But $ \{\{1,2\},\{2,3\}\} $ is not an element of $ X_3^n $ because $ \{1,2\}\cap \{2,3\}\ne \phi $. Note that, for $ i>n+1 $, $ X_i^n=\phi $.

Edit:- A way to imagine the set $ X_k^n $ is to consider the set of un-ordered binary trees of height $ k-1 $ and $ 2^{k-1} $ nodes, whose leaves are numbered by a subset of $ \{1,2,...,2^n\} $ so that the leaves have distinct numbers.

For further clarification, as Amritanshu Prasad mentioned this in comment, $ \{\{\{1,5\},\{2,3\}\}, \{\{1,3\},\{2,4\}\}\} $ is not an element of $ X_3^n $ because $ \{1,5,2,3\}\cap \{1,3,2,4\}\ne \phi $.

I am interested in the following directed system of finite dimensional Hilbert spaces:- $ H_i^n $ is the Hilbert space which has an orthonormal basis $ \{e_s\}_{s\in X_{n+2-i}^n} $. Here $ 1\le i\le n+1 $. We have an isometry $ \psi_i:H_i^n\to H_{i+1}^n\otimes H_{i+1}^n$ given by $ \psi_i(e_{\{x,y\}})=\frac{e_x\otimes e_y+e_y\otimes e_x}{\sqrt {2}}$. This gives the following directed system of Hilbert spaces:- $$ H_1^n\to (H_2^n)^{\otimes 2}\to (H_3^n)^{\otimes 4}\to .....\to (H_{n+1}^n)^{\otimes 2^n} $$ where the map $ (H_i^n)^{\otimes 2^{i-1}}\to (H_{i+1}^n)^{\otimes 2^{i}} $ is given by $ \psi_i^{\otimes 2^{i-1}} $. A nice property of this directed system is that the image of $ \sum\limits_{x\in X_{n+1}^n} e_x\in H_1^n $ in $ (H_{n+1}^n)^{\otimes 2^{n}} $ under the map $ H_1^n\to (H_2^n)^{\otimes 2}\to ....\to (H_{n+1}^n)^{\otimes 2^{n}} $ is $ \sum\limits_{\sigma\in S_{2^n}} e_{\sigma(1)}\otimes e_{\sigma(2)}\otimes ....\otimes e_{\sigma(2^n)} $ upto multiplication by a constant which implies that image of $ \sum\limits_{x\in X_{n+1}^n} e_x\in H_1^n $ under the map $ H_1^n\to (H_2^n)^{\otimes 2}\to ....\to (H_{i}^n)^{\otimes 2^{i-1}} $ lies in $ Sym^{2^{i-1}}(H_i^n) $ but its not of the form $ v^{\otimes 2^{i-1}} $ for some $ v\in H_i^n $.

Suppose, $ H^n $ is the Hilbert space which is the direct limit of the above directed system. I am interested in some specific type of isometries from $ H^n $ to $ H^{n+1} $ and the directed system $$ H^1\to H^2\to H^3\to ... $$ arising from these maps.

I am working on something related to this and I think this type of things about the permutation group $ S_{2^n} $ have been studied before. So any reference would be really helpful.

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  • $\begingroup$ Edited the first paragraph to make it clear. $\endgroup$ – DLN Apr 10 '16 at 21:46
  • $\begingroup$ It looks like $X_k^n$ is empty whenever $2^{k-1} > n$. Am I mistaken? $\endgroup$ – S. Carnahan Apr 11 '16 at 3:19
  • $\begingroup$ Just checking if I understand correctly: is $ \{\{\{1,3\},\{2,5\}\},\{\{1,5\},\{2,3\}\}\} $ also an element of $X^4_3$? $\endgroup$ – Amritanshu Prasad Apr 11 '16 at 4:32
  • $\begingroup$ @Amritanshu Prasad, No. because $ \{1,3,2,5\}\cap \{1,5,2,3\}\ne \phi $ $\endgroup$ – DLN Apr 11 '16 at 5:13
  • $\begingroup$ @S. Carnahan, for any $ x\in X_k^n $, $ \phi_k(x) $ has $ 2^{k-1} $ elements and $ \phi_k(x) $ is a subset of $ X_1^n $. So $ X_k^n=\phi $ for $ k>n+1 $. $\endgroup$ – DLN Apr 11 '16 at 5:18
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These spaces are related to $2$-Sylow subgroups of $S_n$. For example, if $n=2^k$, then $X^n_{k+1}$ is the set of $2$-Sylow subgroups of $S_n$. To see this, note that $S_n$ acts transitively on $X^n_{k+1}$, and that the stabilizer of any fixed element of $X^n_k$ is a $2$-Sylow subgroup of $S_n$. I don't really have a reference for this, but it is not too hard to work out the order of the automorphism group, and see that it is $2^{v_2(n!)}$ using the fact that $$ v_2(n!) = \lfloor n/2 \rfloor + \lfloor n/4 \rfloor + \dotsb $$

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    $\begingroup$ Nice observation. I think you meant $ X_{k+1}^k $ is the set of 2-sylow subgroups of $ S_n $. $\endgroup$ – DLN Apr 12 '16 at 6:24
  • $\begingroup$ Also, I guess we can do the same thing for $ p $-sylow subgroups of $ S_{p^n} $ if instead of taking $ 2 $ element subsets, we take $ p $-element subsets. $\endgroup$ – DLN Apr 12 '16 at 6:58
  • $\begingroup$ I am curious to see the context in which these come up; whenever you are ready to share. You can find my contacts through my profile. $\endgroup$ – Amritanshu Prasad Apr 13 '16 at 4:49

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