Question as the title, I am looking for concret examples of infinite-dimensional Lie algebras without semi-infinite structures. A semi-infinite structure is a necessary condition to do semi-infinite cohomology. A reference for semi-infinite structure and semi-infinite cohomology of Lie algebras is the paper ** Semi-infinite cohomology and string theory** by I. B. Frenkel, H. Garland and G. J. Zuckerman in 1986.

For convenience, I give a brief introduction here.

Basic assumption: $\mathfrak g=\oplus_{n\in \mathbb{Z}} \mathfrak g_n$ is a tame $\mathbb{Z}$-graded Lie algebra, i.e., $\dim \mathfrak g_n< \infty$ and $[\mathfrak g_n, \mathfrak g_m]\subset \mathfrak g_{m+n}$ for all $m, n\in \mathbb{Z}$.

Let $\mathfrak n=\oplus_{n>0}\mathfrak g_n$ and $\mathfrak b=\oplus_{n\leq 0}\mathfrak g_n$ with homogeneous basis $\{e_i~|~i\leq 0\}$ and $\{e_i~|~i>0\}$ respectively. Homogeneous means that each $e_i\in \mathfrak g_m$ for some $m\in \mathbb{Z}$. We also require that whenever $e_i\in \mathfrak g_m$, then $e_{i+1}\in \mathfrak g_m$ or $e_{i+1}\in \mathfrak g_{m+1}$. Let $\mathfrak g'=\oplus_{n\in \mathbb{Z}}\mathfrak g'_n$ be the restricted dual of $\mathfrak g$, where $\mathfrak g'_n:=Hom_{\mathbb{C}}(\mathfrak g_{-n}, \mathbb{C})$, and $\{e_i^*~|~i\in \mathbb{Z}\}$ the dual basis of $\mathfrak g'$ such that $\langle e_i^*, e_j\rangle=\delta_{ij}.$

Definition The space of semi-infinite forms on $\mathfrak g$ is the vector space $\Lambda_{\infty}^*\mathfrak g'$ spanned by the monomials $$\omega=e_{i_1}^*\wedge e_{i_2}^*\wedge\cdots $$ of infinite wedge products of $\mathfrak g'$ satisfying that for each $\omega$, there exists an integer $N(\omega)$ such that when $k>N(\omega)$, we have $i_{k+1}=i_k-1.$

The Clifford algebra $Cl(\mathfrak g\oplus \mathfrak g')$ associated to $\mathfrak g\oplus \mathfrak g'$ is the unital associative algebra generated by $\{e_i, e_i^*~|~i\in \mathbb{Z}\}$ satisfying the relations: \begin{align} e_ie_j+e_je_i=e_i^*e_j^*+e_j^*e_i^*=0,\,\ e_i^*e_j+e_je_i^*=\delta_{ij}, \, \forall~ i, j\in \mathbb{Z}. \end{align} The space $\Lambda_{\infty}^*\mathfrak g'$ is an irreducible $Cl(\mathfrak g\oplus \mathfrak g')$-module with the actions given by \begin{align*} \varepsilon(e_{i_0}^*)\cdot e_{i_1}^*\wedge e_{i_2}^*\wedge\cdots &=e_{i_0}^*\wedge e_{i_1}^*\wedge e_{i_2}^*\wedge\cdots , \\ \iota(e_{i_o})\cdot e_{i_1}^*\wedge e_{i_2}^*\wedge\cdots &=\sum_{k\geq 1}(-1)^{k-1}\langle e_{i_0}, e_{i_k}^*\rangle e_{i_1}^*\wedge e_{i_2}^*\wedge\cdots \wedge \hat{e}_{i_k}^*\wedge \cdots, \end{align*} where the hat over $e_{i_k}^*$ means ommiting this term.

We want to define a $\mathfrak g$-action on $\Lambda_{\infty}^*\mathfrak g'$ (at the moment we just call it an action but not necessarily a Lie algebra action.)

For $x\in \mathfrak g_n$ with $n\neq 0$, we denote by $\rho(x)$, the natural action on $\Lambda_{\infty}^*\mathfrak g'$, \begin{align*} \rho(x)\cdot e_{i_1}^*\wedge e_{i_2}^*\wedge\cdots =\sum_{k\geq 1}e_{i_1}^*\wedge \cdots \wedge ad^*~x(e_{i_k}^*) \wedge\cdots, \end{align*} where $ad^*$ is the coadjoint action of $\mathfrak g$ on $\mathfrak g'$. The above sum is finite hence well-defined. It is easy to verify the following relations (considered as operators on $\Lambda_{\infty}^*\mathfrak g'$): $~\forall~ x, y\in \mathfrak g, y'\in \mathfrak g',$ \begin{equation} [\rho(x), \iota(y)]=\iota(ad~ x(y)), \,\ [\rho(x), \varepsilon(y')]=\varepsilon(ad^*~x(y')). ~~ (\divideontimes) \end{equation}

For elements in $\mathfrak g_0$, we could not use the above definition because it may occur as an infinite sum. We choose a special monomial (there are other choices) $$\omega_0:=e_0\wedge e_{-1}\wedge e_{-2}\wedge \cdots,$$ which could be considered as the maximal form on $\mathfrak b$. It is characterized up to scalar by the property that $\omega_0$ is killed by $\iota(x)$ for all $x\in \mathfrak n$ and $\varepsilon(x')$ for all $x'\in \mathfrak b^*.$

Choose $\beta\in \mathfrak g'_0$ and consider it as a function on $\mathfrak g$. Then for $x\in \mathfrak g_0$, we define its action on $\omega_0$ as $\beta(x)$ and then extend to an action on $\Lambda_{\infty}^*\mathfrak g'$ by requiring the relations $(\divideontimes)$. This can be done because $\Lambda_{\infty}^*\mathfrak g'$ is irreducible and generated by $\omega_0$ as a module of the Clifford algebra $Cl(\mathfrak g\oplus \mathfrak g')$.

To give an explicit expression of $\rho(x)$, we introduce the normal ordering of two operators, \begin{align*} :\iota(e_i)\varepsilon(e_i^*):=\begin{cases} \iota(e_i)\varepsilon(e_i^*),~~ i\leq 0, \\ -\varepsilon(e_i^*)\iota(e_i), ~~ i>0. \end{cases} \end{align*}

Then for any $x\in \mathfrak g$, the following operator acts well on $\Lambda_{\infty}^*\mathfrak g'$, \begin{align}\label{rho} \rho^{\beta}(x)&=\sum_{i\in \mathbb{Z}}:\varepsilon(ad^*x(e_i^*))\iota(e_i):+\beta(x) ~~(\divideontimes\divideontimes) \end{align} and it realizes the action of $x$ on $\Lambda_{\infty}^*\mathfrak g'$ and satisfies the relations $(\divideontimes)$.

Let $$\gamma^{\beta}(x, y)=[\rho^{\beta}(x), \rho^{\beta}(y)]-\rho^{\beta}([x, y]).$$

It can be proved that $\gamma^{\beta}(\cdot, \cdot)$ is a 2-cocycle of $\mathfrak g$ satisfying $\gamma(\mathfrak g_m, \mathfrak g_n)=0$ whenever $m+n\neq 0$.

Definition We call $\mathfrak g$ admits a semi-infinite structure through $\rho^{\beta}$ as defined in $(\divideontimes\divideontimes)$ if the 2-cocycle $\gamma^{\beta}(\cdot, \cdot)\equiv 0$. And we call $\mathfrak g$ admits a semi-infinite structure if $\mathfrak g$ admits a semi-infinite structure through $\rho^{\beta}$ for some $\beta\in \mathfrak g'_0$.

It is obvious that $\mathfrak g$ admits a semi-infinite structure through $\rho^{\beta}$ if and only if the $\mathfrak g$-module $\Lambda_{\infty}^*\mathfrak g'$ is a Lie algebra module under the action $\rho^{\beta}(x)$.

Example When $H^2(\mathfrak g)=0$, every 2-cocycle is a coboundary hence $\mathfrak g$ admits a semi-infinite structure. For example, the affine Kac-Moody algebras and the Virasoro algebra. If $\mathfrak g$ is abelian, it always admits a semi-infinite structure.

Question: Are there examples of tame $\mathbb{Z}$-graded Lie algebras without semi-infinite structures?

  • 1
    $\begingroup$ Please write a definition of "semi-infinite structure" in the question. $\endgroup$
    – S. Carnahan
    Sep 8, 2016 at 14:36
  • $\begingroup$ There's now a restricted access to a definition... Here's a free access definition: p3 of arxiv.org/abs/q-alg/9704020 $\endgroup$
    – YCor
    Sep 8, 2016 at 17:28
  • 1
    $\begingroup$ I haven't fully understood the definition, but part of it is that the Lie algebra has a grading $\bigoplus_{n\in\mathbf{Z}}\mathfrak{g}_n$ with each $\mathfrak{g}_n$ finite-dimensional. Of course it implies countable dimension, but even in this case this is certainly a strong restriction. I guess we can provide (infinite countable)-dimensional Lie algebra (say, complex) with no grading at all in $\mathbf{Z}$ except the trivial one ($\mathfrak{g}_0=\mathfrak{g})$. $\endgroup$
    – YCor
    Sep 8, 2016 at 17:34
  • $\begingroup$ If you have a Lie algebra with a "grading with holes", in the sense that, for instance, $\mathfrak{g}_i=0$ while $\mathfrak{g}_{i-1}$ and $\mathfrak{g}_{i+1}$ are both nonzero, then you can't choose a basis with your requirements. I think I can produce (with some efforts, as in my post) example of Lie algebra with tame gradings, all of whose tame gradings have holes. This would provide counterexamples to your revised question, but I'm afraid you would modify again the question as this is probably not what you want? $\endgroup$
    – YCor
    Sep 18, 2016 at 9:39
  • $\begingroup$ To support my previous comment: there exists a 6-dimensional Lie algebra with a grading with the weights $(1,2,3,4,5,7)$, and all of whose gradings are conjugate to this one, in the sense that for every other $\mathbf{Z}$-grading $\mathfrak{g}=\bigoplus \mathfrak{g}'_i$ with $\mathfrak{g}\neq\mathfrak{g}'_0$, there exists $n\in\mathbf{Z}$ and an automorphism $\phi$ of $\mathfrak{g}$ such that $\mathfrak{g}_i=\phi(\mathfrak{g}'_{ni})$. $\endgroup$
    – YCor
    Sep 18, 2016 at 9:41

1 Answer 1


Edit: the following answers the question before it was edited. The question was to find an infinite-dimensional Lie algebra with no semi-infinite structure. Since the latter involves a $\mathbf{Z}$-grading with finite-dimensional homogeneous components, I answer it below by providing a countable-dimensional Lie algebra admitting no such grading. (In the new version of the question, there is the additional request that the Lie algebra admits such a grading.)

In the sequel, I assume that the ground field is any of zero characteristic, although this is probably not essential.

As I mentioned in a comment, this means that $\mathfrak{g}$ can be endowed with a Lie algebra $\mathbf{Z}$-grading $\mathfrak{g}=\bigoplus_{n\in\mathbf{Z}}\mathfrak{g}_n$ with all $\mathfrak{g}_n$ finite-dimensional and satisfying some complicated extra-data. We can forget those extra data and find Lie algebra having no such grading at all, to answer your question.

First, trivial examples are Lie algebras of uncountable dimension. So I'll consider countable dimension, which is a reasonable restriction. I'll construct an infinite-dimensional Lie algebra with no nonzero grading at all (every Lie algebra can be endowed with the zero grading $\mathfrak{g}=\mathfrak{g}_0$). (If the field is algebraically closed, recall that a finite-dimensional Lie algebra admits no nonzero grading if and only if it is characteristically nilpotent, in the sense that its derivation algebra is nilpotent.)

The construction of $\mathfrak{g}$ below has 2 building blocks: (a) a classical infinite-dimensional "filiform" Lie algebra (b) a 7-dimensional characteristically nilpotent Lie algebra

(a) Consider the 2-generated, infinite-dimensional Lie algebra $\mathfrak{f}$ with basis $(f_i)_{i\ge 1}$. Its nonzero brackets are $[f_i,f_j]=(j-i)f_{i+j}$. ($f_i$ can be represented as $X^{i+1}D$ where $D$ is the derivation operator on the polynomial ring $K[X]$).

(b) Consider the Lie algebra $\mathfrak{e}$ with basis $(e_1,\dots,e_7)$ and nonzero brackets $12=4,14=5,23=24=15=6,13=26=54=7$ (where $ij=k$ stands for $[e_i,e_j]=e_k$). It is borrowed from the classification of 7-dimensional nilpotent Lie algebras: it is denoted as $\mathfrak{g}_{7,0,8}$ in Magnin's list, and has the property that all its derivations are nilpotent ("characteristically nilpotent").

Let $\mathfrak{a}$ be the 2-dimensional abelian Lie algebra with basis $(a_1,a_2)$. We consider homomorphisms $\mathfrak{f}\stackrel{q}\to\mathfrak{a}\stackrel{p}\leftarrow\mathfrak{e}$ given by $f_i,e_{i+1}\mapsto a_i$, $i=1,2$, $f_j\mapsto 0$, $j\ge 3$, $e_1,e_4,\dots,e_7\mapsto 0$.

Finally, $\mathfrak{g}$ can be defined as a subalgebra of the direct product $\mathfrak{f}\times\mathfrak{e}$, namely the fibre product $$\mathfrak{g}=\{(f,e)\in \mathfrak{f}\times\mathfrak{e}:q(f)=p(e)\}.$$ Denoting $g_1=(f_1,e_2)$, $g_2=(f_2,e_3)$, $\mathfrak{g}$ has the basis $$(e_1,g_2,g_3,e_4,\dots,e_7,f_i:i\ge 3),$$ where by abuse of notation we write $e_i$ for $(0,e_i)$, $f_i$ for $(f_i,0)$. Then in this basis, the nonzero brackets are given as $$[g_1,g_2]=f_3+e_6,\quad e_4=[e_1,g_1],\quad e_5=[e_1,e_4],$$ $$e_6=[g_1,e_4]=[e_1,e_5],\quad e_7=[e_1,g_2]=[g_1,e_6]=[e_5,e_4],$$ $$[g_i,f_j]=(j-i)f_{i+j},\quad [f_i,f_j]=(j-i)f_{i+j}.$$ (We could have defined $\mathfrak{g}$ directly this way but this longer construction directly shows, among others, that it satisfies the Jacobi identity.)

For any Lie algebra $\mathfrak{h}$, consider the lower central series given by $\mathfrak{h}^1=\mathfrak{h}$, $\mathfrak{h}^{i+1}=[\mathfrak{h},\mathfrak{h}^i]$. Write $\mathfrak{h}(n)$ for the quotient $\mathfrak{h}/\mathfrak{h}^{n+1}$.

Note that $\mathfrak{f}^2$, which has the basis $(f_i)_{i\ge 3}$, is an ideal in $\mathfrak{g}$: it is the kernel of the (surjective) projection $\mathfrak{g}\to\mathfrak{e}$. Consider a derivation of $\mathfrak{g}$ mapping $\mathfrak{f}^2$ into itself. It induces a derivation on the quotient $\mathfrak{e}$, which is nilpotent. In particular it induces a nilpotent operator on the 3-dimensional abelianization $\mathfrak{g}(1)=\mathfrak{e}(1)$. Therefore, the induced derivation of $\mathfrak{g}(n)$ is nilpotent for all $n$.

We are going to check ($*$) that the 8-dimensional nilpotent Lie algebra $\mathfrak{h}=\mathfrak{g}/\mathfrak{f}^3$ is characteristically nilpotent, ($**$), that $\mathfrak{f}^3$ (which has the basis $(f_i)_{i\ge 4}$) is a characteristic ideal (it is stable under every derivation).

This granted, we can conclude: consider a nonzero grading of $\mathfrak{g}$. Since $\mathfrak{g}^i$ is a graded ideal, it induces a grading on the quotient $\mathfrak{g}(i)$ for all $i$. Also note that the self-operator $D$ acting my multiplication by $i$ on $\mathfrak{g}_i$ is a derivation, and ($**$) thus implies that it passes to the quotient $\mathfrak{g}/\mathfrak{f}^3$, on which $D$ is nilpotent by ($*$); in particular it induces a nilpotent operator on the abelianization $\mathfrak{g}/\mathfrak{g}^2$. Now on a nilpotent Lie algebra, a derivation which is nilpotent on the abelianization is nilpotent. It follows that $D$ is nilpotent on $\mathfrak{g}(n)$ for all $n$, but since it is also diagonalizable, we deduce that $D$ is zero on $\mathfrak{g}(n)$ for all $n$. This means that $D$ takes values in $\mathfrak{g}^{n+1}$ for all $n$. Since the intersection $\bigcap_n\mathfrak{g}^n$ is $0$ (that is, $\mathfrak{g}$ is residually nilpotent), we deduce $D=0$, which means that the grading is zero.

Now let us prove ($*$) (taking for granted that its 7-dimensional quotient $\mathfrak{g}/\mathfrak{f}^2\simeq\mathfrak{e}$ is characteristically nilpotent). It is enough to show that grading on $\mathfrak{h}$ is zero. Fix such a grading; its center is a graded ideal, with basis $(f_3,e_7)$, and $\mathfrak{h}^5$, with basis $(e_7)$, is also graded. So the center contains a homogeneous element of the form $f_3+te_7$. In the quotient by this element, we have $[g_1,g_2]=f_3+e_6=e_6-te_7$, so we get a 7-dimensional Lie algebra with the same law $[\cdot,\cdot]'$ as $\mathfrak{e}$ except $[e_2,e_3]=e_6-te_7$; changing the basis replacing $e_3$ with $e_3+te_6$ yields the same law as $\mathfrak{e}$. Thus the graded quotient $\mathfrak{h}/(f_3+te_7)$ is characteristically nilpotent. In particular the abelianization of $\mathfrak{h}$ has degree zero, which implies that $\mathfrak{h}$ has degree zero, that is, $\mathfrak{h}$ is characteristically nilpotent.

Finally, let us check ($**$). Since $\mathfrak{e}^6=0$, we have $\mathfrak{g}^6\subset\mathfrak{f}$, and more precisely we see that $\mathfrak{g}^6$ has the basis $(f_j)_{j\ge 7}$. Then every derivation of $\mathfrak{g}$ stabilizes $\mathfrak{g}^6$, and therefore stabilizes the centralizer of $\mathfrak{g}^6$, as well at is double centralizer $\mathfrak{s}$. The centralizer of $\mathfrak{g}^6$, is equal to the ideal with basis $(e_1,e_4,e_5,e_6,e_7)$. Its centralizer, hence the double centralizer of $\mathfrak{g}^6$, is equal to $\mathfrak{f}^2\oplus\mathfrak{e}^5$ (where $\mathfrak{e}^5$ is 1-dimensional with basis $e_7$). Finally, we have $[\mathfrak{g},\mathfrak{s}]=\mathfrak{f}^3$.


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