Generators and relations for (generalised) Cartan type superalgebras

TL;DR

This paper introduces a new construction method for Cartan superalgebras, specifically for $W(n)$, using Dynkin diagrams similar to those of contragredient Lie superalgebras but with added generators and relations.

Contribution

It provides a novel presentation of $W(n)$, expanding the understanding of the structure of Cartan superalgebras through Dynkin diagram-based generators and relations.

Findings

New construction of $W(n)$ from Dynkin diagrams

Extended generators and relations for Cartan superalgebras

Enhanced understanding of superalgebra structures

Abstract

In Kac's classification of finite-dimensional Lie superalgebras, the contragredient ones can be constructed from Dynkin diagrams similar to those of the simple finite-dimensional Lie algebras, but with additional types of nodes. For example, $A(n-1,0) = \mathfrak{sl}(1|n)$ can be constructed by adding a "gray" node to the Dynkin diagram of $A_{n-1} = \mathfrak{sl}(n)$, corresponding to an odd null root. The Cartan superalgebras constitute a different class, where the simplest example is $W(n)$, the derivation algebra of the Grassmann algebra on $n$ generators. Here we present a novel construction of $W(n)$, from the same Dynkin diagram as $A(n-1,0)$, but with additional generators and relations.

Equations

$$[x,y]$$

$$[x,[y,z]]$$

$$d(xy)$$

$$[c,d]=c\text{vbox}\circ d-(-1{)}^{|c||d|}d\text{vbox}\circ c,$$

$${\theta }^{a_1}\cdots {\theta }^{a_p}={\theta }^{[a_1}\cdots {\theta }^{a_p]}.$$

$$K^{a_1\cdots a_p}{}_b={\theta }^{a_1}\cdots {\theta }^{a_p}\frac{\partial }{\partial {\theta }^b}$$

$$K^{a_1\cdots a_p}{}_b:{\theta }^{c_1}\cdots {\theta }^{c_q}\mapsto q{\delta }_b{}^{[c_1}{\theta }^{|a_1}\cdots {\theta }^{a_p|}{\theta }^{c_2}\cdots {\theta }^{c_q]}.$$

$$\displaystyle\begin{array}[]{c|c}\text{level}&\text{basis}\\ \hline\cr 1&K_{a}\\ 0&K^{a}{}_{b}\\ -1&K^{ab}{}_{c}\\ \vdots&\vdots\\ -\,n+1&K^{a_{1}\cdots a_{n}}{}_{b}\end{array}$$

$$B_{ij}$$

$$M_{(0)}$$

$$[h_a,e_b]$$

$$(\mathrm{ad}{e}_a{)}^{1-B_{ab}}(e_b)$$

$$B_{ab}=\left(\begin{array}{cccccc}0& -1& 0& \cdots & 0& 0\\ -1& 2& -1& \cdots & 0& 0\\ 0& -1& 2& \cdots & 0& 0\\ ⋮& ⋮& ⋮& \ddots & ⋮& ⋮\\ 0& 0& 0& \cdots & 2& -1\\ 0& 0& 0& \cdots & -1& 2\end{array}\right)$$

$$A(n-1,0)={\mathcal{B}}_{-1}\oplus {\mathcal{B}}_0\oplus {\mathcal{B}}_1.$$

$$\displaystyle\begin{array}[]{c|c|c}\text{generators}&\text{level}&\text{basis}\\ \hline\cr e_{0}&1&E_{a}\\ e_{i},f_{i},h_{a}&0&G^{a}{}_{b}\\ f_{0}&-1&F^{a}\end{array}$$

$$[G^a{}_b,G^c{}_d]$$

$$[G^a{}_b,F^c]$$

$$e_0$$

$$e_i$$

$$M_{(0)}$$

$$[e_0,f_{0a}]$$

$$[e_a,[e_a,e_{0b}]]=[f_a,[f_a,f_{0b}]]=0,$$

$$i,j=2,3,\dots ,r\Rightarrow [e_i,[f_j,f_{0a}]]={\delta }_{ij}{B}_{aj}{f}_{0j},$$

$$[f_{0a},f_{0b}]$$

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Full text

This talk, given by JP at “The 32nd International Colloquium on Group Theoretical Methods in Physics (Group32)” in Prague, 9–13 July, 2018, is based on [1], where more details and references can be found.

In attempts to understand the origin of the duality symmetries appearing in supergravity theories, the Lie algebras $\mathfrak{g}$ describing the symmetries have been extended to infinite-dimensional Lie superalgebras. These extensions include Borcherds superalgebras [2], here denoted by $\mathscr{B}(\mathfrak{g})$, as well as tensor hierarchy algebras [3], here denoted by $W(\mathfrak{g})$ and $S(\mathfrak{g})$.

The construction of tensor hierarchy algebras in [3] was only applicable for finite-dimensional $\mathfrak{g}$ and thus in particular not to the cases $\mathfrak{g}=E_{r}$ for $r\geq 9$. Here we solve this problem by a new construction with generators and relations. We focus on the case $\mathfrak{g}=A_{n-1}$ since $W(\mathfrak{g})$ and $S(\mathfrak{g})$ then turn out to be finite-dimensional and well known as Cartan type superalgebras. In this sense, the general tensor hierarchy algebras $W(\mathfrak{g})$ and $S(\mathfrak{g})$ are generalised Cartan type superalgebras.

We consider algebras over an algebraically closed field $\mathbb{K}$ of characteristic zero. A superalgebra $G$ is an algebra with a $\mathbb{Z}_{2}$-grading, which means that it can be decomposed into a direct sum $G=G_{(0)}\oplus G_{(1)}$ of an even subalgebra $G_{(0)}$ and an odd subspace $G_{(1)}$, such that $G_{(i)}G_{(j)}=G_{(i+j)}$ where $i,j\in\mathbb{Z}_{2}$. (This means that $G_{(1)}$ does not close under the product and is thus not a subalgebra.) In a Lie superalgebra, the product is a bracket that satisfies the identities

[TABLE]

where $|x|=0$ if $x\in G_{(0)}$ and $|x|=1$ if $x\in G_{(1)}$.

A $\mathbb{Z}$-grading of the Lie superalgebra $G$ is a decomposition of $G$ into a direct sum of subspaces $G_{i}$ for all integers $i$, called levels, such that $[G_{i},G_{j}]\subseteq G_{i+j}$. The $\mathbb{Z}$-grading is said to be consistent if $G_{i}\subseteq G_{(i)}$ for all levels $i\in\mathbb{Z}$, that is, if odd elements appear at odd levels and even elements at even levels.

One way to obtain a Lie superalgebra if we only have a superalgebra $G$ to start with (not necessarily a Lie superalgebra) is to consider all even and odd derivations of it. Let $|d|$ be equal to either 1 or 0. A linear map $d:G\to G$ satisfying

[TABLE]

is an odd derivation if $|d|=1$ and an even derivation if $|d|=0$. All even and odd derivations of $G$ span a vector space which, together with the commutator as the bracket,

[TABLE]

forms a Lie superalgebra. This is the derivation algebra of $G$, denoted ${\rm der}\,G$.

The Grassmann algebra $\Lambda(n)$ is a basic example of a superalgebra that is not a Lie superalgebra. It is the associative superalgebra generated by $n$ odd elements $\theta^{0},\theta^{1},\ldots,\theta^{n-1}$, modulo the relations $\theta^{a}\theta^{b}=-\theta^{b}\theta^{a}$, where $a,b=0,1,\ldots,n-1$. It is spanned by monomials $\theta^{a_{1}}\cdots\theta^{a_{p}}$, where $0\leq p\leq n$, which are fully antisymmetric in the upper indices,

[TABLE]

The derivation algebra of $\Lambda(n)$ has a basis consisting of elements

[TABLE]

acting on a monomial $\theta^{c_{1}}\cdots\theta^{c_{q}}$ by a contraction,

[TABLE]

This Lie superalgebra ${\rm der}\,\Lambda(n)$ is also denoted by $W(n)$. It is easy to see that it has a consistent $\mathbb{Z}$-grading where the subspace at level $-p+1$ has a basis of elements $K^{a_{1}\cdots a_{p}}{}_{b}$ which are fully antisymmetric in the upper indices, and thus there are no elements at level $-p+1$ for $p>n$ (or $p<0$). Negative and positive levels are here reversed compared to the usual conventions.

[TABLE]

In the classification of simple finite-dimensional Lie superalgebras, $W(n)$ appear as Lie superalgebras of Cartan type. Such Lie superalgebras are distinguished from the classical Lie superalgebras, which are further divided into basic and strange ones [4].

The basic Lie superalgebras are finite-dimensional cases of contragredient Lie superalgebras [4], which means that they can be constructed from a (generalised) Cartan matrix, or from a Dynkin diagram encoding the same information as the matrix. For a general contragredient Lie superalgebra, the only condition on the Cartan matrix is that it be a square matrix $B_{ab}$ with values in $\mathbb{K}$, where the index set labelling rows and columns is $\mathbb{Z}_{2}$-graded, that is, the disjoint union of an odd and an even subset. Here we restrict to the cases where $B_{ab}$ ($a,b=0,1,2\ldots,r$) is obtained by adding a row and a column to the Cartan matrix $A_{ij}$ ($i,j=1,2\ldots,r$) of a Kac–Moody algebra $\mathfrak{g}$ of rank $r$, such that

[TABLE]

and such that $\{0\}$ and $\{1,2,\ldots,r\}$ are the odd and even subsets, respectively, of the index set $\{0,1,2,\ldots,r\}$. Furthermore, for simplicity we assume that $A_{ij}$ is symmetric (implying that $B_{ab}$ is symmetric as well) and that both $A_{ij}$ and $B_{ab}$ are non-degenerate. To this Cartan matrix we associate a $\mathbb{Z}_{2}$-graded set $M=M_{(0)}\cup M_{(1)}$ of generators,

[TABLE]

where $i=1,2,\ldots,r$ and $a=0,1,2,\ldots,r$. Let $\mathscr{B}(\mathfrak{g})$ be the Lie superalgebra generated by the set $M$ modulo the Chevalley–Serre relations

[TABLE]

[TABLE]

Then $\mathscr{B}(\mathfrak{g})$ is the contragredient Lie superalgebra constructed from the Cartan matrix $B_{ab}$. With the restrictions on $B_{ab}$ here, $\mathscr{B}(\mathfrak{g})$ is not only a contragredient Lie superalgebra but also a Borcherds superalgebra [5, 6].

Let us now apply the above construction to the case of the (finite-dimensional) Kac–Moody algebra $\mathfrak{g}=A_{n-1}$ (thus $r=n-1$) with a Cartan matrix $A_{ij}$ where row and column 1 correspond to one of the end nodes in the Dynkin diagram. Then we get the Cartan matrix $B_{ab}$ of $\mathscr{B}(\mathfrak{g})$ given in (18). We can associate a Dynkin diagram to it, given in Figure 1, where row and column 0 correspond to the “gray” node.

[TABLE]

The resulting Lie superalgebra $\mathscr{B}(A_{n-1})$ is $A(n-1,0)$, one of the basic Lie superalgebras in the classification [4].

If we put $e_{0}$ and $f_{0}$ at level 1 and level $-1$, respectively, and all the other generators at level 0, then we get a consistent $\mathbb{Z}$-grading of $\mathscr{B}(\mathfrak{g})$. In the case $\mathfrak{g}=A_{n-1}$ there are no more levels, this $\mathbb{Z}$-grading of $\mathscr{B}(A_{n-1})$ is a 3-grading:

[TABLE]

The subalgebra $\mathscr{B}_{0}$ is $\mathfrak{sl}(n)\oplus\mathbb{K}=\mathfrak{gl}(n)$, and the basis elements can thus be written as $\mathfrak{gl}(n)$ tensors:

[TABLE]

The commutation relations are

[TABLE]

[TABLE]

where $G=\sum_{a=0}^{n-1}G^{a}{}_{a}$. With the identifications

[TABLE]

[TABLE]

the commutation relations (25) follow from the Chevalley–Serre relations (17).

Let us now compare the basis (24) of $A(n-1,0)$ to the basis (14) of $W(n)$. Level 1 and 0 have the same index structure in $W(n)$ as in $A(n-1,0)$. If we consider level $-1$ in $W(n)$, the tensors can be decomposed into a traceless part and the trace, obtained by contracting the lower index with one of the upper indices. If we take the subalgebra of $W(n)$ generated by level 1 and only the traceless part of level $-1$, then we get another Lie superalgebra of Cartan type, denoted $S(n)$, with traceless tensors all the way down to level $-n+2$. If we instead take the subalgebra generated by level 1 and only the trace part of level $-1$, then there will be no lower levels, and we get $A(n-1,0)$. Thus $A(n-1,0)$ is a subalgebra of $W(n)$, which can be constructed from the generators (16) and the relations (17). The question arises whether we can obtain not only this subalgebra, but the whole of $W(n)$ by extending the set of generators and relations.

We extend the set $M=M_{(0)}\cup M{}_{(1)}$ of generators to $M^{\prime}=M_{(0)}\cup M^{\prime}{}_{(1)}$, where $f_{0}$ is replaced by $r$ generators $f_{0a}$.

[TABLE]

Henceforth, whenever $f_{0a}$ appears we assume $a=0,2,3,\ldots,r$ (with $r=n-1$ in the case of $\mathfrak{g}=A_{n-1}$), and whenever $f_{a}$ appears we assume $a\neq 0$. As we will see, the new generator $f_{00}$ corresponds to the old $f_{0}$. Identifying them with each other, $M^{\prime}$ is indeed an extension of $M$.

We let $\widetilde{W}(\mathfrak{g})$ be the Lie superalgebra generated by the set $M^{\prime}$ modulo the relations (17) and the additional relations

[TABLE]

[TABLE]

[TABLE]

where $a=0,1,2,\ldots,r$.

In the same way as for $\mathscr{B}(\mathfrak{g})$ we get a consistent $\mathbb{Z}$-grading of $\widetilde{W}(\mathfrak{g})$ if we put $e_{0}$ at level 1, all $f_{0a}$ at level $-1$, and all the other generators at level 0. The sum of the subspaces at level $\pm 1$ and 0 constitute the local part of $\widetilde{W}(\mathfrak{g})$. We then define $W(\mathfrak{g})$ as $W(\mathfrak{g})=\widetilde{W}(\mathfrak{g})/J$, where $J$ is the maximal ideal of $\widetilde{W}(\mathfrak{g})$ intersecting the local part trivially.

The following theorem summarises the main results of [1] (where the proof can be found).

Theorem. The Lie superalgebra ${W}(A_{n-1})=\widetilde{W}(A_{n-1})/J$ is isomorphic to $W(n)$. The ideal $J$ of $\widetilde{W}(A_{n-1})$ is generated by the relations

[TABLE]

*where $i,j=3,\ldots,n-1$. Thus $W(n)$ has generators (27) and relations (17), (28) and (29).

*By removing $h_{0}$ and $f_{00}$ from the set $M^{\prime}$ of generators we get a subalgebra $S(\mathfrak{g})$ of $W(\mathfrak{g})$. In the case $\mathfrak{g}=A_{n-1}$ we have $S(\mathfrak{g})=S(n)$. In general, for finite-dimensional $\mathfrak{g}$ this definition of $S(\mathfrak{g})$ agrees with the definition of the corresponding tensor hierarchy algebra in [3]. In cases other than $\mathfrak{g}=A_{n-1}$ we do not know whether the relations (29) generate the whole ideal $J$ or if additional relations are needed.

We conclude this talk with an overview of the cases where $\mathfrak{g}$ belongs to the $A$, $D$ or $E$ series of Kac–Moody algebras, with node 1 being the node to which another node is connected when going to the next algebra in the series. Another Lie superalgebra of Cartan type, $H(2r)$, appears as $S(D_{r})$.

[TABLE]

In applications to extended geometry, the $A$, $D$ and $E$ cases above correspond to ordinary, double and exceptional geometry, respectively [7]. In cases where so-called ancillary transformations are absent, the Borcherds superalgebras can be used to derive expressions for the generalised diffeomorphisms [8] and furthermore an $L_{\infty}$ algebra encoding their gauge structure [9, 10]. When ancillary transformations are present it seems that the Borcherds superalgebra needs to be replaced by a tensor hierarchy algebra [11, 12].