guide:D3a6003619

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Let us go back now to plain easiness, and discuss some classification results, following the old paper , and then the more recent paper of Tarrago-Weber . In order to cut from the complexity, we must impose an extra axiom, and we will use here:

Theorem

For an easy group [math]G=(G_N)[/math], coming from a category of partitions [math]D\subset P[/math], the following conditions are equivalent:

• [math]G_{N-1}=G_N\cap U_{N-1}[/math], via the embedding [math]U_{N-1}\subset U_N[/math] given by [math]u\to diag(u,1)[/math].
• [math]G_{N-1}=G_N\cap U_{N-1}[/math], via the [math]N[/math] possible diagonal embeddings [math]U_{N-1}\subset U_N[/math].
• [math]D[/math] is stable under the operation which consists in removing blocks.

If these conditions are satisfied, we say that [math]G=(G_N)[/math] is uniform.

Show Proof

We use the general easiness theory explained above, as follows:

[math](1)\iff(2)[/math] This is something standard, coming from the inclusion [math]S_N\subset G_N[/math], which makes everything [math]S_N[/math]-invariant. The result follows as well from the proof of [math](1)\iff(3)[/math] below, which can be converted into a proof of [math](2)\iff(3)[/math], in the obvious way.

[math](1)\iff(3)[/math] Given a subgroup [math]K\subset U_{N-1}[/math], with fundamental representation [math]u[/math], consider the [math]N\times N[/math] matrix [math]v=diag(u,1)[/math]. Our claim is that for any [math]\pi\in P(k)[/math] we have:

[[math]] \xi_\pi\in Fix(v^{\otimes k})\iff\xi_{\pi'}\in Fix(v^{\otimes k'}),\,\forall\pi'\in P(k'),\pi'\subset\pi [[/math]]

In order to prove this, we must study the condition on the left. We have:

[[math]] \begin{eqnarray*} \item[a]i_\pi\in Fix(v^{\otimes k}) &\iff&(v^{\otimes k}\xi_\pi)_{i_1\ldots i_k}=(\xi_\pi)_{i_1\ldots i_k},\forall i\\ &\iff&\sum_j(v^{\otimes k})_{i_1\ldots i_k,j_1\ldots j_k}(\xi_\pi)_{j_1\ldots j_k}=(\xi_\pi)_{i_1\ldots i_k},\forall i\\ &\iff&\sum_j\delta_\pi(j_1,\ldots,j_k)v_{i_1j_1}\ldots v_{i_kj_k}=\delta_\pi(i_1,\ldots,i_k),\forall i \end{eqnarray*} [[/math]]

Now let us recall that our representation has the special form [math]v=diag(u,1)[/math]. We conclude from this that for any index [math]a\in\{1,\ldots,k\}[/math], we must have:

[[math]] i_a=N\implies j_a=N [[/math]]

With this observation in hand, if we denote by [math]i',j'[/math] the multi-indices obtained from [math]i,j[/math] obtained by erasing all the above [math]i_a=j_a=N[/math] values, and by [math]k'\leq k[/math] the common length of these new multi-indices, our condition becomes:

[[math]] \sum_{j'}\delta_\pi(j_1,\ldots,j_k)(v^{\otimes k'})_{i'j'}=\delta_\pi(i_1,\ldots,i_k),\forall i [[/math]]

Here the index [math]j[/math] is by definition obtained from [math]j'[/math] by filling with [math]N[/math] values. In order to finish now, we have two cases, depending on [math]i[/math], as follows:

\underline{Case 1}. Assume that the index set [math]\{a|i_a=N\}[/math] corresponds to a certain subpartition [math]\pi'\subset\pi[/math]. In this case, the [math]N[/math] values will not matter, and our formula becomes:

[[math]] \sum_{j'}\delta_\pi(j'_1,\ldots,j'_{k'})(v^{\otimes k'})_{i'j'}=\delta_\pi(i'_1,\ldots,i'_{k'}) [[/math]]

\underline{Case 2}. Assume now the opposite, namely that the set [math]\{a|i_a=N\}[/math] does not correspond to a subpartition [math]\pi'\subset\pi[/math]. In this case the indices mix, and our formula reads:

[[math]] 0=0 [[/math]]

Thus, we are led to [math]\xi_{\pi'}\in Fix(v^{\otimes k'})[/math], for any subpartition [math]\pi'\subset\pi[/math], as claimed.

Now with this claim in hand, the result follows from Tannakian duality.

■

We can now formulate a first classification result, as follows:

Theorem

The uniform orthogonal easy groups are as follows,

[[math]] \xymatrix@R=50pt@C=50pt{ B_N\ar[r]&O_N\\ S_N\ar[u]\ar[r]&H_N\ar[u]} [[/math]]

and this diagram is an intersection and easy generation diagram.

Show Proof

We know that the quantum groups in the statement are indeed easy and uniform, the corresponding categories of partitions being as follows:

[[math]] \xymatrix@R=50pt@C50pt{ P_{12}\ar[d]&P_2\ar[d]\ar[l]\\ P&P_{even}\ar[l]} [[/math]]

Since this latter diagram is an intersection and generation diagram, we conclude that we have an intersection and easy generation diagram of quantum groups, as stated. Regarding now the classification, consider an arbitrary easy group, as follows:

[[math]] S_N\subset G_N\subset O_N [[/math]]

This group must then come from a category of partitions, as follows:

[[math]] P_2\subset D\subset P [[/math]]

Now if we assume [math]G=(G_N)[/math] to be uniform, this category of partitions [math]D[/math] is uniquely determined by the subset [math]L\subset\mathbb N[/math] consisting of the sizes of the blocks of the partitions in [math]D[/math]. Following , our claim is that the admissible sets are as follows:

• [math]L=\{2\}[/math], producing [math]O_N[/math].
• [math]L=\{1,2\}[/math], producing [math]B_N[/math].
• [math]L=\{2,4,6,\ldots\}[/math], producing [math]H_N[/math].
• [math]L=\{1,2,3,\ldots\}[/math], producing [math]S_N[/math].

Indeed, in one sense, this follows from our easiness results for [math]O_N,B_N,H_N,S_N[/math]. In the other sense now, assume that [math]L\subset\mathbb N[/math] is such that the set [math]P_L[/math] consisting of partitions whose sizes of the blocks belong to [math]L[/math] is a category of partitions. We know from the axioms of the categories of partitions that the semicircle [math]\cap[/math] must be in the category, so we have [math]2\in L[/math]. Our claim is that the following conditions must be satisfied as well:

[[math]] k,l\in L,\,k \gt l\implies k-l\in L [[/math]]

[[math]] k\in L,\,k\geq 2\implies 2k-2\in L [[/math]]

Indeed, we will prove that both conditions follow from the axioms of the categories of partitions. Let us denote by [math]b_k\in P(0,k)[/math] the one-block partition, as follows:

[[math]] b_k=\left\{\begin{matrix}\sqcap\hskip-0.7mm \sqcap&\ldots&\sqcap\\ 1\hskip2mm 2&\ldots&k\end{matrix} \right\} [[/math]]

For [math]k \gt l[/math], we can write [math]b_{k-l}[/math] in the following way:

[[math]] b_{k-l}=\left\{\begin{matrix}\sqcap\hskip-0.7mm \sqcap&\ldots&\ldots&\ldots&\ldots&\sqcap\\ 1\hskip2mm 2&\ldots&l&l+1&\ldots&k\\ \sqcup\hskip-0.7mm \sqcup&\ldots&\sqcup&|&\ldots&|\\ &&&1&\ldots&k-l\end{matrix}\right\} [[/math]]

In other words, we have the following formula:

[[math]] b_{k-l}=(b_l^*\otimes |^{\otimes k-l})b_k [[/math]]

Since all the terms of this composition are in [math]P_L[/math], we have [math]b_{k-l}\in P_L[/math], and this proves our first formula. As for the second formula, this can be proved in a similar way, by capping two adjacent [math]k[/math]-blocks with a [math]2[/math]-block, in the middle.

With the above two formulae in hand, we can conclude in the following way:

\underline{Case 1}. Assume [math]1\in L[/math]. By using the first formula with [math]l=1[/math] we get:

[[math]] k\in L\implies k-1\in L [[/math]]

This condition shows that we must have [math]L=\{1,2,\ldots,m\}[/math], for a certain number [math]m\in\{1,2,\ldots,\infty\}[/math]. On the other hand, by using the second formula we get:

[[math]] \begin{eqnarray*} m\in L &\implies&2m-2\in L\\ &\implies&2m-2\leq m\\ &\implies&m\in\{1,2,\infty\} \end{eqnarray*} [[/math]]

The case [math]m=1[/math] being excluded by the condition [math]2\in L[/math], we reach to one of the two sets producing the groups [math]S_N,B_N[/math].

\underline{Case 2}. Assume [math]1\notin L[/math]. By using the first formula with [math]l=2[/math] we get:

[[math]] k\in L\implies k-2\in L [[/math]]

This condition shows that we must have [math]L=\{2,4,\ldots,2p\}[/math], for a certain number [math]p\in\{1,2,\ldots,\infty\}[/math]. On the other hand, by using the second formula we get:

[[math]] \begin{eqnarray*} 2p\in L &\implies&4p-2\in L\\ &\implies&4p-2\leq 2p\\ &\implies&p\in\{1,\infty\} \end{eqnarray*} [[/math]]

Thus [math]L[/math] must be one of the two sets producing [math]O_N,H_N[/math], and we are done.

■

All the above is very nice, but the continuation of the story is more complicated. When lifting the uniformity assumption, the final classification results become more technical, due to the presence of various copies of [math]\mathbb Z_2[/math], that can be added, while keeping the easiness property still true. To be more precise, in the real case, as explained in , we have exactly 6 solutions, which are as follows, with the convention [math]G_N'=G_N\times\mathbb Z_2[/math]:

[[math]] \xymatrix@R=50pt@C=50pt{ B_N\ar[r]&B_N'\ar[r]&O_N\\ S_N\ar[u]\ar[r]&S_N'\ar[u]\ar[r]&H_N\ar[u]} [[/math]]

In the unitary case now, the classification is quite similar, but more complicated, as explained in the paper of Tarrago-Weber . In particular we have:

Theorem

The uniform easy groups which are purely unitary, in the sense that they appear as complexifications of real easy groups, are as follows,

[[math]] \xymatrix@R=50pt@C=50pt{ C_N\ar[r]&U_N\\ S_N\ar[u]\ar[r]&K_N\ar[u]} [[/math]]

and this diagram is an intersection and easy generation diagram.

Show Proof

We know from the above that the groups in the statement are indeed easy and uniform, the corresponding categories of partitions being as follows:

[[math]] \xymatrix@R=50pt@C50pt{ \mathcal P_{12}\ar[d]&\mathcal P_2\ar[d]\ar[l]\\ P&\mathcal P_{even}\ar[l]} [[/math]]

Since this latter diagram is an intersection and generation diagram, we conclude that we have an intersection and easy generation diagram of groups, as stated. As for the uniqueness result, the proof here is similar to the proof from the real case, from Theorem 15.32, by examining the possible sizes of the blocks of the partitions in the category, and doing some direct combinatorics. For details here, we refer to Tarrago-Weber .

■

Finally, let us mention that the easy quantum group formalism can be extended into a “super-easy” group formalism, covering as well the symplectic group [math]Sp_N[/math]. This is something a bit technical, and we refer here to the paper of Collins-\'Sniady .

General references

Banica, Teo (2024). "Linear algebra and group theory". arXiv:2206.09283 [math.CO].