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Mutations and braid group actions

When people say SOD they usually mean a longer one in general. We can give a general definition along the lines of the two term one.

Definition. A sequence of $\mathcal A_1,\ldots,\mathcal A_l$ of full triangulated subcategories of $\mathcal T$ is a semi-orthogonal decomposition if

  • for each $1 \leq i \leq l$ we have \(\mathcal A_i \subseteq {}^\perp\langle \mathcal A_1,\ldots,\mathcal A_{i-1} \rangle \cap \langle \mathcal A_{i+1},\ldots,\mathcal A_l \rangle^\perp\)
  • for each object $X$ of $\mathcal T$ there is a diagram

with $A_i \in \mathcal A_i$ for each $i$, the dashed arrows are of homological degree $1$, and the triangles are actual triangles in $\mathcal T$.

Let’s assume that each of the $\mathcal A_i$ are admissible subcategories. Then, we are free to mutate them.

Given a length $l$ SOD $\langle \mathcal A_1, \ldots, \mathcal A_l \rangle$, we have the left mutation at $i$ \(\mathbf{L}_i := \langle \mathcal A_1, \ldots, \mathcal A_{i-1}, \mathcal A_{i+1}^\prime, \mathcal A_i, \mathcal A_{i+2}, \ldots, \mathcal A_l \rangle\) with \(\mathcal A_{i+1}^\prime := {}^\perp\langle \mathcal A_1,\ldots,\mathcal A_{i-1} \rangle \cap \langle \mathcal A_i, \mathcal A_{i+1},\ldots,\mathcal A_l \rangle^\perp\) and right mutation \(\mathbf{R}_i := \langle \mathcal A_1, \ldots, \mathcal A_{i-1}, \mathcal A_{i+1}, \mathcal A_i^{\prime}, \mathcal A_{i+2}, \ldots, \mathcal A_l \rangle\) with \(\mathcal A_i^\prime := {}^\perp\langle \mathcal A_1,\ldots,\mathcal A_{i-1},\mathcal A_{i+1} \rangle \cap \langle \mathcal A_{i+2},\ldots,\mathcal A_l \rangle^\perp\) Each is can be accomplishbed via left/right mutation in the subcategory $\langle \mathcal A_i, \mathcal A_{i+1} \rangle$. Given this, we have already seen that $\mathbf{L}_i^{-1} = \mathbf{R}_i$

Let $\operatorname{SOD}_l(\mathcal T)$ denote the set of semi-orthogonal decompositions of length $l$. Then, we get an action of the free group on ${1,\ldots,l}$ on $\operatorname{SOD}_l(\mathcal T)$ with $j$ acting via $\mathbf{L}_j$.

Proposition. We have the following relations:

  • if $|i - j| > 1$, \(\mathbf{L}_i \mathbf{L}_j = \mathbf{L}_j \mathbf{L}_i\)
  • \[\mathbf{L}_i \mathbf{L}_{i+1} \mathbf{L}_i = \mathbf{L}_{i+1}\mathbf{L}_i \mathbf{L}_{i+1}\]
Proof. (Expand to view)

Since the categories $$ \langle \mathcal A_i, \mathcal A_{i+1} \rangle = \langle \mathcal A_{i+1}^\prime, \mathcal A_i \rangle = \langle \mathcal A_{i+1}, \mathcal A_i^\prime \rangle $$ all coincide, we can compute the left or right orthogonal using any of them when they all lie on the same the term. Assume that $i < j$ for simplicity. Then the $j$-th component of $\mathbf{L}_i \mathbf{L}_j$ is $$ {}^\perp\langle \mathcal A_1,\ldots,\mathcal A_{j-1} \rangle \cap \langle \mathcal A_{j+1},\ldots,\mathcal A_l \rangle^\perp $$ where as the $j$-th component of $\mathbf{L}_j \mathbf{L}_i$ is $$ {}^\perp\langle \mathcal A_1,\ldots,\mathcal A_{j+1}^\prime, \mathcal A_j,\ldots,\mathcal A_{j-1} \rangle \cap \langle \mathcal A_{j+1},\ldots,\mathcal A_l \rangle^\perp $$ Since $$ \langle \mathcal A_1,\ldots,\mathcal A_{j-1} \rangle = \langle \mathcal A_1,\ldots,\mathcal A_{i+1}^\prime, \mathcal A_i,\ldots,\mathcal A_{j-1} \rangle $$ these coincide. Similarly, one can check the other components coincide in the case $|i - j| > 1$. For the other identity, it suffices to consider the case $\langle \mathcal A, \mathcal B, \mathcal C \rangle$. We only need to identity two components as any component is determined by the two others. The third component of both is $\mathcal A$ by definition. The second component of $\mathbf{L}_1 \mathbf{L}_2 \mathbf{L}_1$ is $$ \mathcal B^\prime = \langle \mathcal A, \mathcal C \rangle^\perp $$ Since $$ \langle \mathcal A, \mathcal B \rangle = \mathcal C^\perp $$ and $$ \langle \mathcal A, \mathcal C \rangle^\perp = \mathcal A^\perp \cap \mathcal C^\perp $$ we see that $$ \langle \mathcal A, \mathcal B \rangle \cap \mathcal A^\perp = \mathcal B^\prime $$ The second component of $\mathbf{L}_2 \mathbf{L}_1 \mathbf{L}_2$ is $$ \widetilde{\mathcal B}^{\prime} = {}^\perp \left( \mathcal C^{\prime\prime} \right) \cap \mathcal A^\perp $$ where $$ \mathcal C^{\prime \prime} = \langle \mathcal A, \mathcal B \rangle^\perp $$ So $$ \widetilde{\mathcal B}^{\prime} = {}^\perp \left( \langle \mathcal A, \mathcal B \rangle^\perp \right) \cap \mathcal A^\perp $$ But $$ {}^\perp \left( \langle \mathcal A, \mathcal B \rangle^\perp \right) = \langle \mathcal A, \mathcal B \rangle $$ since we have an SOD $$ \langle \mathcal C^{\prime \prime}, \langle \mathcal A, \mathcal B \rangle \rangle \rangle $$ Thus, $$ \mathcal B^\prime = \widetilde{\mathcal B}^{\prime} $$

The group with the presentation \(B_{n+1} := \langle \sigma_1, \ldots, \sigma_n \mid \sigma_i\sigma_j = \sigma_j\sigma_i \text { if } |i-j| > 1 \text{ and } \sigma_i\sigma_{i+1}\sigma_i = \sigma_{i+1}\sigma_i \sigma_{i+1} \rangle\) has a natural geometric meaning. It is the braid group on $n+1$ strands. This can be described as follows.

Elements of this group are isotopy classes in $\mathbb{R}^3$ of smooth cobordisms between the points $n+1$ points on a line in the plane and $n+1$ points on a parallel line in the plane. Collapsing the points should not know the strands. Composition is concatenation.

For example,

is an element of $B_3$ represented by $s_1 s_2^{-1}$ (reading left to right).

It is straightforward to see that the relations in $B_{n+1}$ hold for the braid group. It is a fact, due to (Emil) Artin, that those generate all the relations.

As a consequence of the presentation, we get an action of $B_l$ the set of semi-orthogonal decompositions of length $l$. Not much is known about this action in general. For example, can we classify the orbits in some concrete way?