Higher spin ${{\mathfrak {s}}}{{\mathfrak {l}}}_2$R-matrix from equivariant (co)homology

Abstract

We compute the rational ${{\mathfrak {s}}}{{\mathfrak {l}}}_2$R-matrix acting in the product of two spin-$\ell \over 2$ (${\ell \in {\mathbb {N}}}$) representations, using a method analogous to the one of Maulik and Okounkov, i.e., by studying the equivariant (co)homology of certain algebraic varieties. These varieties, first considered by Nekrasov and Shatashvili, are typically singular. They may be thought of as the higher spin generalizations of $A_1$ Nakajima quiver varieties (i.e., cotangent bundles of Grassmannians), the latter corresponding to $\ell =1$.

Appendices

Appendix A: ${\mathbb {Z}}_2$ invariance of ${\mathfrak {M}}$ and ${\mathfrak {M}}_1$

In this section, we denote the dual weight by $k^\vee =n\ell -k$.

Lemma 3

There are two isomorphisms:

${\mathfrak {M}}(k, n, \ell )\simeq {\mathfrak {M}}(k^\vee , n, \ell )$ and ${\mathfrak {M}}_1(k, n, \ell )\simeq {\mathfrak {M}}_1(k^\vee , n, \ell )$.

Proof

Remembering that ${{\widetilde{Q}}}: {\mathbb {C}}^k\rightarrow {\mathbb {C}}^n$, we form the matrix ${\widetilde{A}}: {\mathbb {C}}^k\rightarrow {\mathbb {C}}^{\ell n}$ as follows:

$$\begin{aligned} {\widetilde{A}}=\begin{pmatrix}{{\widetilde{Q}}}\\ {{\widetilde{Q}}}\Phi \\ \vdots \\ {{\widetilde{Q}}}\Phi ^{\ell -1} \end{pmatrix} \end{aligned}$$

(62)

(We recall that $\Phi ^\ell =0$.) Earlier, we introduced the stability condition ${{\mathrm {rank}}}\,{\widetilde{A}}=k$. The dimension of the kernel of ${\widetilde{A}}$ is therefore $k^\vee$. We define a matrix ${\widetilde{B}}: {\mathbb {C}}^{k^\vee }\rightarrow {\mathbb {C}}^{\ell n}$ by the equation

$$\begin{aligned} {\widetilde{B}}^T{\widetilde{A}}=0 \end{aligned}$$

(63)

and the non-degeneracy requirement ${\mathrm {rank}}\,{\widetilde{B}}=k^\vee$. The matrix ${\widetilde{A}}$ was defined up to right multiplication by an element of $GL(k, {\mathbb {C}})$, and the matrix ${\widetilde{B}}$ is accordingly defined up to right multiplication by the element of $GL(k^\vee , {\mathbb {C}})$.

Now, due to the structure (62) of the matrix ${\widetilde{A}}$, multiplication by $\Phi$ from the right essentially amounts to shifting the entries of A upwards by n rows, and filling the lower rows with zeros. Denoting such a shift matrix by $S_n$, we write

$$\begin{aligned} {\widetilde{A}}\Phi =S_n {\widetilde{A}}\,. \end{aligned}$$

(64)

Now, multiplying (63) by $\Phi$ from the right, we obtain

$$\begin{aligned} {\widetilde{B}}^T{\widetilde{A}}\Phi ={\widetilde{B}}^T S_n {\widetilde{A}}=0\, \end{aligned}$$

(65)

hence $S_n^T{\widetilde{B}}\in {\mathrm {Ker}}({\widetilde{A}}^T)$. (Note that $S_n^T$ acts on ${\widetilde{B}}$ by shifting by n rows downward.) Since ${\mathrm {Ker}}({\widetilde{A}}^T)$ is spanned by the columns of ${\widetilde{B}}$, we may write

$$\begin{aligned} S_n^T{\widetilde{B}}={\widetilde{B}}\Phi ^{\vee } ,\quad \quad \quad \Phi ^{\vee }\in {\mathrm {End}}({\mathbb {C}}^{k^\vee })\,. \end{aligned}$$

(66)

Since $(S_n)^\ell =0$, we find that $(\Phi ^{\vee })^\ell =0$. Moreover, we can rephrase (66) as the requirement that ${\widetilde{B}}$ has the form

$$\begin{aligned} {\widetilde{B}}=\begin{pmatrix}{{\widetilde{Q}}}^\vee (\Phi ^\vee )^{\ell -1} \\ {{\widetilde{Q}}}^\vee (\Phi ^\vee )^{\ell -2}\\ \vdots \\ {{\widetilde{Q}}}^\vee \end{pmatrix}. \end{aligned}$$

(67)

Therefore, we have defined a map

$$\begin{aligned} \{{{\widetilde{Q}}}, \Phi \}^s/GL(k, {\mathbb {C}}) \leftrightarrow \{{{\widetilde{Q}}}^\vee , \Phi ^\vee \}^s/GL(k^\vee , {\mathbb {C}}). \end{aligned}$$

(68)

We have therefore proved the duality of projective retracts ${\mathfrak {P}}(k, n, \ell )\simeq {\mathfrak {P}}(k^\vee , n, \ell )$.

We now wish to incorporate the variable Q in the above discussion. To this end, we form the $\Phi$-tower of Q that we will call $A: {\mathbb {C}}^{\ell n}\rightarrow {\mathbb {C}}^k$, analogously to the way it was done in (62) for the ${{\widetilde{Q}}}$-variable:

$$\begin{aligned} A=\{\Phi ^{\ell -1} Q , \ldots , \Phi Q , Q\}\,. \end{aligned}$$

(69)

In this case, the NS-equation $\sum \nolimits _{i+j=\ell -1}\,\Phi ^{i}\,Q\,{\widetilde{Q}}\,\Phi ^{j}=0$ may be put in the form

$$\begin{aligned} A{\widetilde{A}}=0\,, \end{aligned}$$

(70)

which is the NS-generalization of the equation $Q{\widetilde{Q}}=0$ in the hyper-Kähler case. To complete the duality, we impose the following condition:

$$\begin{aligned} ({\widetilde{A}}A)^T={\widetilde{B}}B\,. \end{aligned}$$

(71)

Note that in both sides we have matrices of size $\ell n\times \ell n$. A simple consistency check of the above equations is obtained by multiplying from the left by ${\widetilde{A}}^T$. The l.h.s. is then $({\widetilde{A}}A {\widetilde{A}})^T=0$ by (70), and the r.h.s. is ${\widetilde{A}}^T{\widetilde{B}}B=0$ by (63).

Since ${\widetilde{B}}$ has already been defined as (67) from (63), one can view Eq. (71) as an equation for B. Multiplying (71) by ${\widetilde{B}}$ from the right and using (63), we get ${\widetilde{B}}B {\widetilde{B}}=0$. Since $B{\widetilde{B}}\in {\mathrm {End}}({\mathbb {C}}^{k^\vee })$ and ${\mathrm {rank}}({\widetilde{B}})=k^\vee$, it follows that

$$\begin{aligned} B {\widetilde{B}}=0\,, \end{aligned}$$

(72)

which is an analogue of (70).

It remains to show that B, just like A, may be written as a $\Phi$-tower of a basic element $Q^\vee$. To this end, we multiply $({\widetilde{A}}A)^T$ by $S_n^T$ from the right and use (64): $({\widetilde{A}}A)^T S_n^T=(S_n{\widetilde{A}}A)^T=({\widetilde{A}}\Phi A)^T=({\widetilde{A}}A S_n)^T=S_n^T({\widetilde{A}}A )^T$. Here, we have used $\Phi A=A S_n$. By (71) this implies ${\widetilde{B}}B S_n^T=S_n^T {\widetilde{B}}B$. Using (66), ${\widetilde{B}}(B S_n^T-\Phi ^{\vee } B)=0$. Since ${\widetilde{B}}: {\mathbb {C}}^{k^\vee }\rightarrow {\mathbb {C}}^{\ell n}$ and ${\mathrm {rank}}({\widetilde{B}})=k^\vee$, we get $B S_n^T-\Phi ^{\vee } B=0$, hence

$$\begin{aligned} B=\{ Q^\vee , \ldots , (\Phi ^\vee )^{\ell -2} Q^\vee , (\Phi ^\vee )^{\ell -1} Q^\vee \}\,. \end{aligned}$$

(73)

Thus, ${\mathfrak {M}}(k, n, \ell )\simeq {\mathfrak {M}}(k^\vee , n, \ell )$.

In order to prove the isomorphism ${\mathfrak {M}}_1(k, n, \ell )\simeq {\mathfrak {M}}_1(k^\vee , n, \ell )$, we will show that, under the map (68), we have ${\mathfrak {M}}^o_1(k, n, \ell )\leftrightarrow {\mathfrak {M}}^o_1(k^\vee , n, \ell )$.

First we set, as earlier, $k=q\ell +r$, where $0\le r <\ell$. If $\Phi \in {\mathcal {O}}_1(k, \ell )$, then $\Phi$ has the Jordan form

. This may be reformulated as follows. Consider the truncated matrix ${\widetilde{A}}_j: {\mathbb {C}}^k\rightarrow {\mathbb {C}}^{(\ell -j+1)n}$, defined as

$$\begin{aligned} {\widetilde{A}}_j=\begin{pmatrix}{{\widetilde{Q}}}\Phi ^{j-1}\\ \vdots \\ {{\widetilde{Q}}}\Phi ^{\ell -1} \end{pmatrix}. \end{aligned}$$

(74)

Since ${{\widetilde{Q}}}$ contains the generators of all Jordan towers, ${\mathrm {Span}}(\text {rows of }{\widetilde{A}}_j)\simeq {\mathrm {Im}}(\Phi ^{j-1})$. From the Jordan form of $\Phi$, it follows that

$$\begin{aligned}&{\mathrm {rank}}\,{\widetilde{A}}_j={\mathrm {dim}}\;{\mathrm {Im}}(\Phi ^{j-1})=(\ell -j+1)q \quad \text {for}\quad j>r\,, \end{aligned}$$

(75)

$$\begin{aligned}&{\mathrm {rank}}\,{\widetilde{A}}_j={\mathrm {dim}}\;{\mathrm {Im}} (\Phi ^{j-1})=(\ell -j+1)q+(r-j+1) \quad \text {for}\quad j\le r\,. \end{aligned}$$

(76)

For the dimension of the null-space, we therefore get ${\mathrm {dim}}\;{\mathrm {Ker}}\,{\widetilde{A}}_j^T=(\ell -j+1)n-{\mathrm {rank}}\,{\widetilde{A}}_j$. From the fact that the matrices ${\widetilde{A}}_j$ are nested inside each other, it follows that ${\mathrm {Ker}}\,{\widetilde{A}}_\ell ^T \subset {\mathrm {Ker}}\,{\widetilde{A}}_{\ell -1}^T\subset \cdots \subset {\mathrm {Ker}}\,{\widetilde{A}}^T$.

Next, we consider the truncated ${\widetilde{B}}$-matrices ${\widetilde{B}}_j: {\mathbb {C}}^{k^\vee }\rightarrow {\mathbb {C}}^{n j}$

$$\begin{aligned} {\widetilde{B}}_j=\begin{pmatrix}{{\widetilde{Q}}}^\vee (\Phi ^\vee )^{j-1} \\ \vdots \\ {{\widetilde{Q}}}^\vee \end{pmatrix}. \end{aligned}$$

(77)

The vectors spanning ${\mathrm {Ker}}\,{\widetilde{A}}_j^T$ enter in some of the columns of ${\widetilde{B}}_{\ell -j+1}$, and accordingly the vectors spanning ${\mathrm {Ker}}\,{\widetilde{A}}_{j-1}^T$ enter in some of the columns of the matrix ${\widetilde{B}}_{\ell -(j-1)+1}$, which is obtained from ${\widetilde{B}}_{\ell -j+1}$ by adding the additional rows ${{\widetilde{Q}}}^\vee (\Phi ^\vee )^{\ell -j+1}$ on top. Moreover, we have the embedding ${\mathrm {Ker}}\,{\widetilde{A}}_j^T\subset {\mathrm {Ker}}\,{\widetilde{A}}_{j-1}^T$. The vectors spanning ${\mathrm {Ker}}\,{\widetilde{A}}_{j-1}^T \setminus {\mathrm {Ker}}\,{\widetilde{A}}_j^T$ remain linearly independent even upon truncation to the “upper block” ${{\widetilde{Q}}}^\vee (\Phi ^\vee )^{\ell -j+1}$. Therefore,

$$\begin{aligned} {\mathrm {rank}}({{\widetilde{Q}}}^\vee (\Phi ^\vee )^{\ell -j+1})\ge {\mathrm {dim}}({\mathrm {Ker}}\,{\widetilde{A}}_{j-1}^T \setminus {\mathrm {Ker}}\,{\widetilde{A}}_j^T)\,. \end{aligned}$$

(78)

Using

$$\begin{aligned} {\mathrm {dim}}({\mathrm {Ker}}\,{\widetilde{A}}_{1}^T)=\ell (n-q)-r,\quad \quad {\mathrm {dim}}({\mathrm {Ker}}\,{\widetilde{A}}_{2}^T)=(\ell -1)(n-q)-{\mathrm {max}}(r-1,0)\,,\nonumber \\ \end{aligned}$$

(79)

we find from (78) for $j=2$:

$$\begin{aligned} {\mathrm {rank}}({{\widetilde{Q}}}^\vee (\Phi ^\vee )^{\ell -1})\ge n-q+{\mathrm {max}}(r-1,0)-r={\left\{ \begin{array}{ll} n-q\quad \text {if}\quad r=0 \\ n-q-1 \quad \text {if}\quad r>0 \end{array}\right. } \end{aligned}$$

(80)

Since $k^\vee =n\ell -k=(n-q-1)\ell +\ell -r$, there can be at most $n-q-1$ or $n-q$ Jordan towers of height $\ell$ for $r>0$ and $r=0$, respectively. As the number of such maximal towers is measured by ${\mathrm {rank}}({{\widetilde{Q}}}^\vee (\Phi ^\vee )^{\ell -1})$, the inequality (80) shows that this bound is saturated. This implies that $\Phi ^\vee$ has the Jordan structure

, i.e., $\Phi ^\vee \in {\mathcal {O}}_1(k^\vee , \ell )$.

Now we need to show that $Q^\vee {{\widetilde{Q}}}^\vee \in T_{\Phi ^\vee } {\mathcal {O}}_1(k^\vee , \ell )$. Since the defining equations of ${\mathcal {O}}_1(k^\vee , \ell )$ are $(\Phi ^\vee )^\ell =0$ and ${{\,\mathrm{tr}\,}}((\Phi ^\vee )^i)=0$ for all i, we need to show that ${d\over d\epsilon }(\Phi +\epsilon Q^\vee {{\widetilde{Q}}}^\vee )^\ell \big |_{\epsilon =0}=0$ and ${{\,\mathrm{tr}\,}}((\Phi ^\vee )^{i-1}Q^\vee {{\widetilde{Q}}}^\vee )=0$ for all i. The first part follows from (72), if one recalls the definitions (67) and (73). To prove the second part, we look more closely at the relation (71). The $\ell n\times \ell n$-matrix ${\widetilde{A}}A$ may be viewed as an $\ell \times \ell$-matrix with the entries $({\widetilde{A}}A)^T_{ij}={{\widetilde{Q}}}\Phi ^{\ell -1+j-i} Q$, analogously $({\widetilde{B}}B)_{ij}={{\widetilde{Q}}}^\vee (\Phi ^\vee )^{\ell -1+j-i} Q^\vee$. Therefore, (71) is equivalent to ${{\widetilde{Q}}}\Phi ^{\ell -1+j-i} Q={{\widetilde{Q}}}^\vee (\Phi ^\vee )^{\ell -1+j-i} Q^\vee$ for all i, j. Since $(Q, {{\widetilde{Q}}}, \Phi )\in {\mathfrak {M}}^o_1(k, n, \ell )$, ${{\,\mathrm{tr}\,}}(Q{{\widetilde{Q}}}\Phi ^{i})=0$ is satisfied for all i. It then follows that ${{\,\mathrm{tr}\,}}(Q^\vee {{\widetilde{Q}}}^\vee (\Phi ^\vee )^{i})=0$ for all i, as required. Therefore, $Q^\vee {{\widetilde{Q}}}^\vee \in T_\Phi {\mathcal {O}}_1(k^\vee , \ell )$ and $(Q^\vee , {{\widetilde{Q}}}^\vee , \Phi ^\vee )\in {\mathfrak {M}}^o_1(k^\vee , n, \ell )$. $\square$

Appendix B: The inverse matrix $S^{-1}$

To prove the formula (60) for $S^{-1}$, we observe that $S^{-1}S$ is a rational function of z. We will now compute the residues of $(S^{-1}S)_{ij'}$ at the potential poles and show that they vanish.

First, changing $r\rightarrow -r$ in the third factor in the denominator of (59), we simplify the expression for $S_{jj'}$:

$$\begin{aligned} S_{jj'}= \frac{(-1)^{j'-j}{j'\atopwithdelims ()j} \prod _{r=j}^{j'-1}(r\varphi +\varepsilon )}{ \prod _{r=0}^{k-j-1} (r\varphi +\varepsilon +z) \prod _{r=k-j-j',\;r\ne k-2j}^{k-j}(r\varphi +z) } \end{aligned}$$

Using this, as well as (60), we find

$$\begin{aligned} \sum \limits _{j=i}^{j=j'}\,S^{-1}_{ij} S_{jj'}=\prod \limits _{r=i}^{j'-1}(r \varphi +\epsilon )\,\sum \limits _{j\in {\mathbb {Z}}}\,{j' \atopwithdelims ()j} {j\atopwithdelims ()i}\frac{(-1)^{j'-j}((k-2j)\varphi +z)}{\prod _{r=k-j-j'}^{k-j-i} (r\varphi +z)} \end{aligned}$$

(81)

We have extended the sum over j to the integers, as the binomial coefficients suppress all terms outside of $i\le j\le j'$. Clearly, the potential poles of the r.h.s. are at $z=-n \varphi , n\in {\mathbb {Z}}$. Note that there are exactly $j'-i+1$ factors in the denominator of each term: Setting $r=k-j-j'+r'$, these factors correspond to $r'=0\ldots j'-i$. The ${\tilde{r}}$th factor vanishes at $z=-n \varphi$ when $k-j-j'+{\tilde{r}}=n$, i.e., this happens in the jth term in the sum, where $j=k-j'+{\tilde{r}}-n$. The residue at this point is the sum of $j'-i+1$ terms:

$$\begin{aligned} \underset{z=-n \varphi }{\mathrm {res}}\left( S^{-1}S\right) _{ij'}=\prod \limits _{r=i}^{j'-1}(r \varphi +\epsilon )\,\sum \limits _{{\tilde{r}}=0}^{j'-i}{j' \atopwithdelims ()j} {j\atopwithdelims ()i}\frac{(-1)^{j'-j}(k-2j-n)\varphi }{\prod _{r'=0, r'\ne {\tilde{r}}}^{j'-i} ((r'-{\tilde{r}})\varphi )}\big |_{j=k-j'+{\tilde{r}}-n}=\\ =\prod \limits _{r=i}^{j'-1}(r \varphi +\epsilon )\,\frac{(-1)^{k-n}}{\varphi ^{j'-i-1}} \sum \limits _{{\tilde{r}}\in {\mathbb {Z}}}{j' \atopwithdelims ()j} {j\atopwithdelims ()i}\frac{(k-2j-n)}{{\tilde{r}}! (j'-i-{\tilde{r}})!}\big |_{j=k-j'+{\tilde{r}}-n}. \end{aligned}$$

We have again extended the summation to the integers, due to the presence of the factorials in the denominator. The factor $(k-2j-n)\big |_{j=k-j'+{\tilde{r}}-n}$ in the numerator is skew-symmetric under the change ${\tilde{r}}\rightarrow 2j'-(k-n)-{\tilde{r}}$. At the same time, one can check that the product of factorials and binomial coefficients in the sum is symmetric under this change. Therefore, $\underset{z=-n \varphi }{\mathrm {res}}\left( S^{-1}S\right) _{ij'}=0$.

It is obvious from (81) that $(S^{-1}S)_{ij'}$ vanishes for $z\rightarrow \infty$ if $j'-i>0$ and approaches 1 if $j'=i$. We conclude that $(S^{-1}S)_{ij'}=\delta _{ij'}$.