Generating Empty Branes from the Euler Sequence

The following example will illustrate how one arrives at the formula for generating empty branes from the Euler sequence on the Grassmannian, and how one finds the corresponding Young tableau \(\lambda \) for a chosen \(\mu \).

A useful example is that of \(\mu =(2,1)\) for \(K_\). The Schur functor on the Euler sequence is

$$\begin L_ \left[ S \rightarrow V \rightarrow Q \right] . \end$$

(A.1)

Given that the sequence in the brackets is exact, so too is

$$\begin L_ \left[ S \rightarrow V \right] \rightarrow L_ \left[ Q \right] . \end$$

(A.2)

The second object involves those pesky sheaves constructed from Q, so the following steps are taken to convert them to sheaves constructed from \(S^\vee \).

First, take a power of Q equal to the total number of boxes in \(\mu \). In this case,

$$\begin Q \otimes Q \otimes Q \cong L_ Q \oplus 2 L_ Q \oplus L_ Q. \end$$

(A.3)

Due to

$$\begin \wedge ^k E \cong \wedge ^r E \otimes \wedge ^ E^\vee \end$$

(A.4)

the power of Q is also isomorphic to

$$\begin Q^&\cong \left( \det Q \otimes \wedge ^2 Q^\vee \right) ^ \end$$

(A.5)

$$\begin&\cong (\det Q )^ \otimes \left( L_ Q^\vee \oplus 2L_Q^\vee \oplus L_ Q^\vee \right) . \end$$

(A.6)

By equating the coefficients and/or ranks of the bundles, one can infer the relation

$$\begin L_Q \cong (\det Q )^ \otimes L_ Q^\vee . \end$$

(A.7)

The \(\lambda \) here is then the Young tableau (3, 2, 1) for \(\mu =(2,1)\).

Occasionally, the relation can be simplified further. For this example, \(L_ Q^\vee \) can be decomposed into

$$\begin L_ Q^\vee&\cong L_ Q^\vee \otimes L_ Q^\vee \end$$

(A.8)

$$\begin&\cong \det Q^\vee \otimes L_ Q^\vee , \end$$

(A.9)

reducing Eq. (A.7) to

$$\begin L_Q \cong (\det Q )^ \otimes L_ Q^\vee . \end$$

(A.10)

The relations \(\det Q \cong \det S^\vee \) and \(Q \cong S \rightarrow V\) (from the Euler sequence) can then be utilised to replace sheaves of Q for S:

$$\begin (\det S^\vee )^ \otimes L_} (V^\vee \rightarrow S^\vee ). \end$$

(A.11)

The Schur functor on the Euler sequence for this example then becomes

$$\begin L_(S \rightarrow V) \rightarrow (\det S^\vee )^ \otimes L_} (V^\vee \rightarrow S^\vee ). \end$$

(A.12)

Examples of Generating Empty Branes for Monodromy Actions

In this section, we collect two examples, namely the cases \(K_\) and \(K_\), to illustrate how shifting between windows in the \(\xi \gg 1\) phase can be performed, using only empty branes constructed from exact sequences of the form \(L_E\), where E is the Euler sequence.

1.1 \(K_\)

Consider an arbitrary B-brane \(\mathcal \) with charge configuration belonging to the window \(\omega _\) in Fig. 9. To undergo monodromy for a loop around \(\theta =-\pi \), we must window shift through \(\omega _\) (shown also in Fig. 9). The charge needed to be replaced is on the \(n=0\) shifted diagonal. Specifically, we need to replace the module \(\mathcal _\). In a first step, we need to do it using an empty brane in the \(\xi \ll -1\) phase, namely (as pointed out in Sect. 5.4.1):

$$\begin \mathcal ^_ = \mathcal _ \rightarrow \mathcal _. \end$$

(B.1)

Upon a twist by \(\mathcal _\), it serves our purpose: taking multiple cones with \(\mathcal \) will restrict \(\mathcal \) to \(\omega _\).

Finally we have to return to \(\omega _\), but using empty branes in the \(\xi \gg 1\) phase. So, we need to ‘replace back’ \(\mathcal _\) by branes with charges in \(\omega _\). This can be done by taking cones with (\(\mathcal _\) twists of) the UV lift of:

$$\begin \pi ^*L_ E, \end$$

(B.2)

since this empty brane will have a copy of \(\mathcal _\) on one end. Indeed, it is equivalent to

(B.3)

However, the module in bold in (B.3), namely does not fit the window \(\omega _\) (also upon twisting by \(\mathcal _\)). Therefore we need to keep taking cones by other empty branes, in this case, a logical candidate will be the UV lift the sequence \(\pi ^*(\wedge ^2 V \otimes L_ E)\), i.e.:

(B.4)

Afterwards there are no more branes left with charges outside \(\omega _\). This means, in order to shift between the window \(\omega _\) to \(\omega _\) in the \(\xi \gg 1\) phase, we only need to take multiple cones by the UV lifts of \(\pi ^*(L_ E\otimes S^)\) and \(\pi ^*(L_ E)\), i.e. a finite number of empty branes.

Now consider a B-brane \(\mathcal \) grade-restricted to \(\omega _\), but we undergo monodromy through \(\theta =-2\pi \). Then we need to go through the window \(\omega _\). The charges needed to be replaced are on the \(n=1\) shifted diagonal and corresponds to the charges of \(\mathcal _\).

This can be done in the \(\xi \ll -1\) phase by using the empty brane:

$$\begin \mathcal ^_ = \mathcal _ \rightarrow \mathcal _, \end$$

(B.5)

upon twisting by \(\mathcal _\). For the window shift, from \(\omega _\) back to \(\omega _\) in the positive phase, we proceed likewise. First we take cones by (the \(\mathcal _\) twist of) the UV lift of

$$\begin \pi ^*L_ E, \end$$

(B.6)

which is equivalent to the sequence

This time, we marked two modules in bold in (B.7) that do not belong to \(\omega _\). There are two more empty branes required to remove these. On the one hand, we have the exact sequence:

$$\begin \pi ^*\left( V \otimes L_ E \right) , \end$$

(B.8)

which is equivalent to

$$\begin \pi ^*\left[ \mathbf } \rightarrow S^ \rightarrow \mathcal ^ \rightarrow S^ \rightarrow S^ \right] ^, \end$$

(B.9)

and on the other hand we have the exact sequence

$$\begin&\pi ^*\left[ V \otimes \textrm^2 S \otimes L_ E\right] ^\vee \nonumber \\&=\pi ^*\left[ S^ \otimes \left( \mathbf } \rightarrow S^ \rightarrow \mathcal ^ \rightarrow S^ \rightarrow S^ \right) \right] ^, \end$$

(B.10)

or, equivalently

$$\begin \pi ^*\left[ S^ \rightarrow S^ \rightarrow S^ \rightarrow S^ \rightarrow \mathbf } \right] ^. \end$$

(B.11)

Afterwards there are no more bundles left, with charges outside \(\omega _\). This means, in order to shift between the window \(\omega _\) to \(\omega _\) in the \(\xi \gg 1\) phase, we only need to take multiple cones with the UV lifts of \(\pi ^*\left( L_ E\otimes S^\right) \), \(\pi ^*\left( L_ E\otimes S^\right) \) and \(\pi ^* \left[ \left( \textrm^2 S \otimes L_ E\right) ^\otimes S^\right] \).

1.2 \(K_\)

In this section we perform the monodromy for \(K_\). We consider a B-brane \(\mathcal \) grade-restricted to the window \(\omega '_\), as described in (3.57). This window is shown in the top left diagram of Fig. 5. We will start by taking a loop around \(\theta =-2\pi \) through th window \(\omega _\). Note we can choose \(\omega '_\) or any other valid window configuration we want. In any case, in order to shift from \(\omega '_\) to \(\omega _\) we need to exchange the module \(\mathcal _\) by \(\mathcal _\). Their weights belong to the \(n=0\) shifted diagonal so, in order to map \(\mathcal \) to \(\omega _\), using an empty brane in the \(\xi \ll -1\) phase, we need to take successive cones with:

$$\begin \mathcal ^_=\mathcal _ \rightarrow \mathcal _, \end$$

(B.12)

upon twisting it by \(\mathcal _\). Finally we need to return to \(\omega '_\) using an empty brane in the \(\xi \gg 1\) phase. We can do this by taking cones with the UV lift of (upon twisting it by \(\mathcal _\))

$$\begin \pi ^*L_ E. \end$$

(B.13)

This is equivalent to

The module \(\mathcal _\), in bold, (or more precisely \(\mathcal _\otimes \mathcal _=\mathcal _\)) do not belong to \(\omega '_\) but still maps \(\mathcal \) to another valid window inside \(\hat_\). If we insist on return to the configuration \(\omega '_\) we just can take cones with the empty brane from the exact sequence \(L_ E\) (twisted by \(\mathcal _\)):

(B.14)

Therefore we conclude that, in the \(\xi \gg 1\) phase, we only need to take cones with the UV lifts of \(\pi ^*\left( L_E\otimes S^\right) \) and, possibly \(\pi ^*\left( L_E\otimes S^\right) \), if we insist to return to \(\omega '_\).

Now we consider the case of encircling the singularities at \(\theta =-3\pi \). We again consider \(\mathcal \in \omega '_\). In order to map it to \(\omega _\) (for any valid choice of \(\omega _\)) we need to exchange the module \(\mathcal _\) by \(\mathcal _\). They belong to the \(n=1\) shifted diagonal. Then such a map, in the \(\xi \ll -1\) phase, will be implemented by successive cones by \(\mathcal ^_\otimes \mathcal _\) where

$$\begin \mathcal ^_=\mathcal _ \rightarrow \mathcal _. \end$$

(B.15)

For the map \(\omega _\rightarrow \omega '_\) in the \(\xi \gg 1\) phase we can start by considering the UV lift of \(\pi ^*\left( L_ E\otimes S^\right) \), where the UV lift of \(\pi ^*L_ E\) is explicitly given by

The two modules in bold have different meaning. The module \(\mathcal _\) does not belong to any valid subwindow of \(\hat_\), therefore it must be removed. We can do it by taking cones with the UV twist of (twisted by \(\mathcal _\))

$$\begin \pi ^*[S^\otimes L_E]^, \end$$

(B.18)

where

$$\begin L_ E = \mathbf } \rightarrow \underline} \rightarrow S^ \rightarrow \mathcal ^ \rightarrow S^ \rightarrow S^. \end$$

(B.19)

To other module in bold in (B.17), i.e. the module \(\mathcal _\) needs only to be removed if we insist on keeping the original shape of the window \(\omega '_\). As in the case of monodromy around \(\theta =-2\pi \), we can just use the empty brane from the exact sequence \(L_ E\otimes S^\) for this purpose. In summary, monodromy around \(\theta =-3\pi \) requires us to take cones, in the \(\xi \gg 1\) phase, with the UV lifts of \(\pi ^*\left( L_E\otimes S^\right) \), \(\pi ^*\left[ (\textrm^S\otimes L_E)^\otimes S^\right] \) and, possibly \(\pi ^*\left( L_E\otimes S^\right) \), if we insist to return to \(\omega '_\).