Restricted Kac modules of Hamiltonian Lie superalgebras of odd type

Abstract

This paper aims to describe the restricted Kac modules of restricted Hamiltonian Lie superalgebras of odd type over an algebraically closed field of characteristic $p>3$. In particular, a sufficient and necessary condition for the restricted Kac modules to be irreducible is given in terms of typical weights.

Appendix

I. Let

$$\begin{aligned} \phi (x_{i})=\left\{ \begin{array}{ll} \sum _{j=1}^{n}a_{ji}x_{j}&{} \quad \hbox {if} \; i\le n \\ \sum _{j=n+1}^{2n}a_{ji}x_{j}&{} \quad \hbox {if}\; i>n, \end{array} \right. \end{aligned}$$

where $a_{ji}\in \mathbb {F}.$ If $\phi =\mathrm {diag}(t,\ldots , t)\in \mathfrak {J},$ then

$$\begin{aligned} f_{\phi }\left( \sum _{i=1}^{2n}x_{i}\partial _{i}\right)&= \sum _{i=1}^{2n}\phi (x_{i})f_{\phi }(\partial _{i})=\sum _{i=1}^{2n}tx_{i}t^{-1}\partial _{i}= \sum _{i=1}^{2n}x_{i}\partial _{i} \end{aligned}$$

(5.1)

and

$$\begin{aligned} f_{\phi }(\mathrm {De}_{f})=t^{\mathrm {deg}f-2}\mathrm {De}_{f} \quad \;\hbox {for}\; f\in \mathcal {O}(n). \end{aligned}$$

(5.2)

If $\phi \in \mathrm {SP}(n, \mathbb {F}),$ then for $1 \le i, j\le n,$ we have $\sum _{k=1}^{n}a_{ik}a_{j^{\prime }{k^{\prime }}}= \delta _{ij}.$ Then

$$\begin{aligned} f_{\phi }\left( \sum _{i=1}^{2n}x_{i}\partial _{i}\right)&= \sum _{i=1}^{2n}\phi (x_{i})f_{\phi }(\partial _{i})\nonumber \\&= \sum _{i=1}^{n}\sum _{j=1}^{n}a_{ji}x_{j}\sum _{k=1}^{n}a_{{k^{\prime }}{i^{\prime }}}\partial _{k}+\sum _{i=n+1}^{2n}\sum _{j=n+1}^{2n}a_{ji}x_{j}\sum _{k=n+1}^{2n} a_{{k^{\prime }}{i^{\prime }}}\partial _{k}\nonumber \\&= \sum _{j,k=1}^{n}\left( \sum _{i=1}^{n}a_{ji}a_{{k^{\prime }}{i^{\prime }}}\right) x_{j}\partial _{k}+\sum _{j,k=n+1}^{2n}\left( \sum _{i=n+1}^{2n}a_{ji}a_{k^{\prime }}{i^{\prime }}\right) x_{j}\partial _{k}\nonumber \\&= \sum _{j,k=1}^{2n}\delta _{jk}x_{j}\partial _{k}=\sum _{i=1}^{2n}x_{i}\partial _{i} \end{aligned}$$

(5.3)

and

$$\begin{aligned} f_{\phi }(\mathrm {De}_{f})=\mathrm {De}_{\phi (f)} \quad \hbox {for}\; f\in \mathcal {O}(n). \end{aligned}$$

(5.4)

Eqs. (5.2) and (5.4) imply that

$$\begin{aligned} f_{\phi }([\mathrm {De}_{f}, \mathrm {De}_{g}])=[f_{\phi }(\mathrm {De}_{f}),f_{\phi }(\mathrm {De}_{g})] \quad \hbox {for}\; f, g\in \mathcal {O}(n). \end{aligned}$$

Using Eqs. (5.1–5.4) and the fact that $\phi$ is a $\mathbb {Z}$-homogeneous automorphism of $\mathcal {O}(n)$, we have

$$\begin{aligned} f_{\phi }\left( \left[ \sum _{i=1}^{n}x_{i}\partial _{i}, \mathrm {De}_{f}\right] \right) =\left[ f_{\phi }\left( \sum _{i=1}^{n}x_{i}\partial _{i}\right) ,f_{\phi }(\mathrm {De}_{f})\right] \quad \hbox {for}\; f\in \mathcal {O}(n). \end{aligned}$$

Therefore, $f_{\phi }\in \mathrm {Aut}\left( \overline{\mathfrak {le}}(n)\right) .$

II. As in [7], we have

$$\begin{aligned} X(\mathfrak {T})/ p X(\mathfrak {T})\cong \Lambda _{\overline{\mathfrak {le}}(n)}. \end{aligned}$$

Then for $a\in \mathbf {u}(\overline{\mathfrak {le}}(n))$ and a highest weight vector $\upsilon _{\lambda }$ of $L_{\overline{\mathfrak {le}}(n)}^{\mathfrak {b}_{0}}(\lambda )$ with respect to $\mathfrak {b}_{0}$, we can define

$$\begin{aligned} \overline{t}(a\otimes v_{\lambda })=\mathrm {Ad}(\overline{t})(a)\otimes \lambda (\overline{t})(v_{\lambda }). \end{aligned}$$

Clearly, $I_{\overline{\mathfrak {le}}(n)}(\lambda )_{\bar{i}},\,i=0, 1$, is a $\mathfrak {T}$-module and

$$\begin{aligned} \overline{t}(a\cdot v)=\mathrm {Ad}(\overline{t})(a)\overline{t}(v) \quad \hbox {for}\;\overline{t}\in \mathfrak {T}, a\in \mathbf {u}(\overline{\mathfrak {le}}(n)), v\in I_{\overline{\mathfrak {le}}(n)}(\lambda ). \end{aligned}$$

We claim that the action of $\mathfrak {T}$ on $\overline{\mathfrak {le}}(n)$ coincides with that of $\bar{\mathfrak {h}}.$ For

$$\begin{aligned} \overline{t}=\mathrm {diag}(tt_{1},\ldots ,tt_{n},tt_{1}^{-1},\ldots ,tt_{n}^{-1})\in \mathfrak {T},\;x^{(\underline{r})}\in \mathcal {O}(n)\quad \hbox {and}\quad h\in \bar{\mathfrak {h}}, \end{aligned}$$

we can check the following equations:

$$\begin{aligned} \overline{t}(\mathrm {De}_{x^{(\underline{r})}})&= t^{\mathrm {deg}x^{(\underline{r})}-2} t_{1}^{r_{1}}\ldots t_{n}^{r_{n}}t_{1}^{-r_{n+1}}\ldots t_{n}^{-r_{2n}}\mathrm {De}_{x^{(\underline{r})}}\\&= \left( \sum _{i=1}^{n}(r_{i^\prime }-r_{i})\Lambda _{i}+(\mathrm {deg}x^{(\underline{r})}-2)\Lambda _{n+1}\right) (\overline{t})\mathrm {De}_{x^{(\underline{r})}}, \end{aligned}$$

$$\begin{aligned}&\displaystyle [h,\mathrm {De}_{x^{(\underline{r})}}]=\left( \sum _{i=1}^{n}(r_{i^\prime }-r_{i}) \varepsilon _{i}+(\mathrm {deg}x^{(\underline{r})}-2)\delta \right) (h)\mathrm {De}_{x^{(\underline{r})}},&\\&\displaystyle \overline{t}\left( \sum _{i=1}^{2n}x_{i}\partial _{i}\right) =\sum _{i=1}^{2n}x_{i}\partial _{i}=0(\overline{t})\sum _{i=1}^{2n}x_{i}\partial _{i},&\\&\displaystyle \left[ h,\sum _{i=1}^{2n}x_{i}\partial _{i}\right] =0=0(h)\sum _{i=1}^{2n}x_{i}\partial _{i}.&\end{aligned}$$

Summarizing, the action of $\mathfrak {T}$ on $\overline{\mathfrak {le}}(n)$ coincides with that on $\bar{\mathfrak {h}}.$ Then $I_{\overline{\mathfrak {le}}(n)}(\lambda )$ is a $\left( \mathbf {u}(\overline{\mathfrak {le}}(n)),\mathfrak {T}\right)$-module. Similarly, $L_{\overline{\mathfrak {le}}(n)}^{\mathfrak {b}_{i}}(\lambda )$ is also a $\left( \mathbf {u}(\overline{\mathfrak {le}}(n)),\mathfrak {T}\right)$-module, $0 \!\le \! i\!\le \! 2n$.