Let $\mathfrak{g}$ be a solvable Lie algebra over $\mathbb{C}$ and $\lambda\in(\mathfrak{g}/[\mathfrak{g},\mathfrak{g}])^*$ be a character of $\mathfrak{g}$. I'm interested in calculating homology for $\mathbb{C}_\lambda$, the one-dimensional $\mathfrak{g}$-module with character $\lambda$.

I have calculated homology manually for the 2-dimensional algebra $\mathfrak{g}=\langle x,y \rangle$ with relation $[x,y]=y$. The thing that surprised me is that homology is nontrivial only for $\lambda(x)=0$ or $1$.

In general, I conjecture is that the homology is nontrivial, iff $\lambda$ is a weight of the adjoint representation. I can't prove it (or find counterexample). I've tried to find the answer in books, but there is a lack on literature in homology theory for solvable lie algebras.

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    $\begingroup$ Weight of the adjoint rep. in the metabelianization $\endgroup$ – YCor May 6 at 17:27
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    $\begingroup$ How did you get the criterion? $\endgroup$ – Boris Bilich May 6 at 18:05
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    $\begingroup$ I guess that Guichardet's book "Cohomologie des groupes topologiques et des algèbres de Lie" might contain relevant information. Unfortunately it's not easy to find. For instance it probably includes the fact that $H_*(\mathfrak{g},V_\lambda)=0$ for $\mathfrak{g}$ nilpotent and all $\lambda\neq 0$. $\endgroup$ – YCor May 7 at 10:51
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    $\begingroup$ Using this paper numdam.org/article/CM_1984__53_3_347_0.pdf of Brown I think you can easily get sufficient conditions for all homology groups to vanish. The idea I have in mind is that you can localise $U(\mathfrak{g})$ at the maximal Ore set $S$ contained in the complement of the augmentation ideal (described in the paper) and then compute homology by taking $(\mathbb{C}_S\otimes^L_{U(\mathfrak{g})_S} (\mathbb{C}_{\lambda})_S)$. $(\mathbb{C}_{\lambda})_S$ will already be zero except for on an explicitly parameterised set of $\lambda$. $\endgroup$ – Simon Wadsley May 7 at 13:58
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    $\begingroup$ If you want more details I can supply them another time. You can find the general idea in section 3 in one of my papers dpmms.cam.ac.uk/~sjw47/Euler.pdf but for another similar context. $\endgroup$ – Simon Wadsley May 7 at 13:59

Let me provide an elementary proof that for $\mathfrak{g}$ nilpotent finite-dimensional and $\lambda\neq 0$ we have $H_*(\mathfrak{g},V_\lambda)=0$. (In general it seems to be particular case of results of Delorme in the 70s, and possibly known earlier.)

Recall that the homology of $V_\lambda$ is the homology of the Chevalley-Eilenberg complex, with $C_p=\Lambda^p\mathfrak{g}$, and $d_p:C_p\to C_{p-1}$ is given by $$x_1\wedge\dots\wedge x_p\mapsto \sum_{i=1}^p(-1)^{i+1}\lambda(x_i)x_1\wedge\dots\wedge \hat{x_i}\wedge\dots\wedge x_p+$$ $$+\sum_{1\le i<j\le p}[x_i,x_j]\wedge x_1\wedge\dots\wedge \hat{x_i}\wedge\dots\wedge \hat{x_j}\wedge\dots\wedge x_p.$$

Let $\mathfrak{z}(\mathfrak{g})$ be the center of $\mathfrak{g}$. Then for all $z\in\mathfrak{z}(\mathfrak{g})\cap [\mathfrak{g},\mathfrak{g}]$ and $u\in\Lambda^{p-1}\mathfrak{g}$ we have $d_p(u\wedge z)=d_{p-1}(u)\wedge z$ (because all remaining terms include $[x_i,z]$ or $\lambda(z)$ which are zero). The induction step is the following:

Lemma. Fix $p\ge 1$. Let $z$ be an element of $\mathfrak{z}(\mathfrak{g})\cap [\mathfrak{g},\mathfrak{g}]$. Suppose that $H_i(\mathfrak{g}/Kz,V_\lambda)$ is zero for $i\in\{p-1,p\}$. Then $H_p(\mathfrak{g},V_\lambda)=0$.

Let $f$ be a cycle in degree $p$. Modulo $z$ it is a $p$-boundary. This means that there exists $g\in\Lambda^{p+1}\mathfrak{g}$ such that $d_{p+1}(g)=f$ in $\lambda^p\mathfrak{g}/Kz$, which means that $d_{p+1}(g)=f+\mu\wedge z$ for some $\mu\in\Lambda^{p-1}\mathfrak{g}$. Since $d_pf=0$, we moreover have $d_{p-1}(\mu)\wedge z=d_p(\mu\wedge z)=0$. This means that $d_{p-1}(\mu)=0$ in $\Lambda^{p-2}(\mathfrak{g}/Kz)$. Hence, there exists $\xi\in\Lambda^{p}\mathfrak{g}$ such that $d_{p}\xi=\mu$ in $\Lambda^{p-1}(\mathfrak{g}/Kz)$. This means that $d_p\xi=\mu+\eta\wedge z$ for some $\eta\in\Lambda^{p-2}\mathfrak{g}$. So $d_{p+1}(\xi\wedge z)=d_p(\xi)\wedge z=\mu\wedge z$ and hence $f$ is a boundary. (If $p=1$, we directly have $d_0\mu=0$ since $d_0=0$ so no need to introduce $\eta$.)

Corollary. $H_*(\mathfrak{g},V_\lambda)=0$ for $\mathfrak{g}$ finite-dimensional nilpotent and $\lambda\neq 0$.

Given that $H_0(\mathfrak{g},V_\lambda)=0$ for every $\lambda\neq 0$, the lemma reduces, by induction, to the case when $\mathfrak{g}$ is abelian. In this case, we choose the basis $(e_0,\dots,e_n)$ with $\lambda(e_0)=1$, $\lambda(e_i)=1$ for $i\ge 1$, and view the Lie algebra as graded in $\mathbf{Z}^n$ with $e_i$ of grade $u_i$ (the canonical basis of $\mathbf{Z}^n$) and $e_0$ of grade $0$, noting that the boundary map preserves the grading. In $\Lambda^i\mathfrak{g}$, we have the grades $u_I=\sum_{i\in I}u_i$ when $I$ is a subset of $\{1,\dots,n\}$ of cardinal $i$ or $i-1$. If $I$ has cardinal $i$, $(\Lambda^i\mathfrak{g})_{u_I}$ has dimension $1$, generated by the $x_I=\wedge_{i\in I}e_i$, which is a $i$-cycle, and is also a boundary, namely of $e_0\wedge x_I$ (up to sign). If $I$ has cardinal $i-1$, $(\Lambda^i\mathfrak{g})_{u_I}$ has dimension $1$, generated by $y_I=e_0\wedge x_I$, which has a nonzero image by $d_i$. This proves the vanishing of all the homology.

  • $\begingroup$ Can this be reworked into a spectral sequence argument? It just has that feel about it... $\endgroup$ – Victor Protsak May 8 at 3:40

Let $K$ be a field. Let $\mathfrak{g}$ be a Lie algebra over $K$. Let $\lambda:\mathfrak{g}\to K$ be a homomorphism. Let $V_\lambda$ be $K$ endowed with the structure of right $\mathfrak{g}$-module given by $\lambda$.

For a representation $V$ of $\mathfrak{g}$, let us say that $\lambda$ is a weight of $V$ is $V_\lambda$ is isomorphic to a subquotient of $V$ as $\mathfrak{g}$-module. This means that for some block-triangulation of the representation, $\lambda$ appears as a diagonal block. Say that $\lambda$ is a strong weight if $V_\lambda$ is isomorphic to a submodule of $V$, i.e., if there exists $y\in V\smallsetminus\{0\}$ such that $x\dot y=\lambda(x)y$ for all $x\in\mathfrak{g}$. For $\mathfrak{g}$ nilpotent and $\dim(V)<\infty$, weight and strong weight are equivalent, but not in general: for instance, for the adjoint representation of the 2-dimensional Lie algebra $\mathfrak{b}$ with basis $u,v$, $[u,v]=v$ and $\lambda(u)=0$ and $\lambda(v)=0$, the weights are $0$ and $\lambda$, but only $\lambda$ is a strong weight.

By definition, we have $H_1(\mathfrak{g},V_\lambda)=\mathrm{Ker}(\lambda)/I_\lambda$, where $I_\lambda\subset\mathfrak{g}$ is the image of $d:\Lambda^2\mathfrak{g}\to\mathfrak{g}$, $x\wedge y\mapsto \lambda(x)y-\lambda(y)x-[x,y]$.

If $\lambda=0$, then $H_1(\mathfrak{g},V_0)$ is the abelianization, and hence is zero iff $\mathfrak{g}$ is perfect (which is the same as $0$ being a strong weight of the coadjoint representation. Note that for $\mathfrak{g}$ finite-dimensional, $0$ is a weight of the adjoint/coadjoint representation iff it is not semisimple (for $\mathfrak{g}$ solvable this means iff $\mathfrak{g}\neq 0$). It is a strong weight of the adjoint representation iff $\mathfrak{g}$ has nontrivial center.

Next I assume $\lambda\neq 0$.

The non-vanishing of $H_1(\mathfrak{g},V_\lambda)$ means that there exists a linear form $f$ on $\mathfrak{g}$, not proportional to $\lambda$, such that $f([x,y])=\lambda(x)y-\lambda(y)x$ for all $x,y$. The latter condition means that $x\mapsto \lambda(x)u+f(x)v$ is a homomorphism into the 2-dimensional Lie algebra $\mathfrak{b}$, and the non-proportionality means that it is nonzero. In other words, the vanishing of $H_1(\mathfrak{g},V_\lambda)$ means that every lift $\mathfrak{g}\to\mathfrak{b}$ of $\lambda$ has a 1-dimensional image.

It is clear that the latter condition only depends on the metabelianization $\mathfrak{g}/\mathfrak{g}^{(2)}$.

Also it's clear that if $\mathfrak{g}$ is nilpotent then $\mathfrak{b}$ is not quotient of $\mathfrak{g}$, so in this case $H_1(\mathfrak{g},V_\lambda)=0$ for all $\lambda\neq 0$.

For $\mathfrak{g}$ finite-dimensional metabelian and $\lambda\neq 0$, let me check that $H_1(\mathfrak{g},V_\lambda)\neq 0$ iff $\lambda$ is weight of the adjoint representation, if and only if $\lambda$ is a strong weight of the adjoint representation.

Proof: the latter equivalence is true for metabelian Lie algebras and $\lambda\neq 0$. Indeed, if $\mathfrak{a}$ is a Cartan subalgebra and $\mathfrak{n}$ is the derived subalgebra, let $\lambda$ be a nonzero weight of $\mathfrak{g}$. So it is a strong weight of $\mathfrak{n}$ on $\mathfrak{n}$. If $y\neq 0$ and $[x,y]=\lambda(x)y$ for all $\in\mathfrak{a}$, then this is also true for $x\in\mathfrak{n}$ (because $\mathfrak{n}$ is abelian), and hence for all $x$ (since $\mathfrak{a}+\mathfrak{n}=\mathfrak{g}$).

If the $H_1(\mathfrak{g},V_\lambda)$ is nonzero, we have a surjection $\mathfrak{g}\to\mathfrak{b}$ projecting to $\lambda$ and by pull-back we deduce that $\lambda$ is a weight of $\mathfrak{g}$.

Conversely, $\lambda$ is a weight of $\mathfrak{g}$, by the above it is a strong weight. So there exists $y$ such that for all $x\in\mathfrak{g}$ we have $[x,y]=\lambda(x)y$. Since $\lambda\neq 0$, necessarily $y\in\mathfrak{g}^{(1)}$, the derived subalgebra, and this can be viewed as a condition on the adjoint representation of $\mathfrak{g}/\mathfrak{g}^{(1)}$ on $\mathfrak{g}^{(1)}$. We can decompose this representation into isotypic components, and kill all other components (for other eigenvalues as $\lambda$), and also kill an invariant hyperplane in the isotypic component of $\lambda$. After doing this, we preserve the property of $\lambda$ being a weight (maybe $y$ is not longer the same), and in addition the derived subalgebra is 1-dimensional. Using a Cartan subalgebra, we see that $\mathfrak{g}$ is then semidirect product of an abelian subalgebra $\mathfrak{a}$ with the 1-dimensional $\mathfrak{g}$, and killing the kernel of $\lambda$ in $\mathfrak{a}$ results in a 2-dimensional algebra. This quotient is precisely a 2-dimensional surjective lift of $\lambda$ onto $\mathfrak{b}$.

Corollary: for $\mathfrak{g}$ finite-dimensional and $\lambda\neq0$
$H_1(\mathfrak{g},V_\lambda)\neq 0$ iff $\lambda$ is a (strong) weight of the adjoint representation of $\mathfrak{g}$ on the metabelianization $\mathfrak{g}/\mathfrak{g}^{(2)}$.

(a) Now here's an example for which $\lambda$ is a weight of the adjoint representation of $\mathfrak{g}$, but not of its metabelianization (so $H_1(\mathfrak{g},V_\lambda)=0$ anyway). Namely, consider a Lie algebra with basis $(s,x,y,z)$ and nonzero brackets $[s,x]=x$, $[s,y]=y$, $[s,z]=2z$, $[x,y]=z$. (This appears a parabolic subalgebra in $\mathfrak{su}(2,1)$ over the reals.) Define $\lambda$ mapping $s$ to 1 and other basis elements to 0. Then $2\lambda$ is a (strong) weight of the adjoint representation (with eigenspace generated by $z$), but not a weight of the 3-dimensional metabelianization (which only has the weights $0$ and $\lambda$), so $H_1(\mathfrak{g},V_{2\lambda})=0$. [Edit: actually $H_*(\mathfrak{g},V_{2\lambda})=0$, see (e) below.]

Edits: here are remarks about $H_0$ and $H_d$ for $d=\dim(\mathfrak{g})$. First, we have $H_0(\mathfrak{g},V_\lambda)=0$ iff $\lambda\neq 0$.

(b) For $\mathfrak{g}$ finite-dimensional, say of dimension $d$, define $\tau_{\mathfrak{g}}(x)=\mathrm{Trace}(\mathrm{ad}(x))$ (for instance, $\mathfrak{g}$ is unimodular iff $\tau_{\mathfrak{g}}= 0$). Then $H_d(\mathfrak{g},V_\lambda)=0$ if and only if $\lambda=\tau_{\mathfrak{g}}$. However, $\tau_{\mathfrak{g}}$ is often not a weight.

(c) For instance, if $\mathfrak{g}$ is 3-dimensional with 1-dimensional abelianization and weights of the adjoint representation $0$, $\lambda$ and $t\lambda$ with $t\notin\{0,-1\}$, then $\tau_{\mathfrak{g}}=(1+t)\lambda$ is not a weight of the adjoint representation although $H_*(\mathfrak{g},V_{\tau_{\mathfrak{g}}})\neq 0$.

(d) Looking at $H_2$ also provides unimodular counterexamples. Namely, consider an $(n+1)$-dimensional Lie algebra with basis $(s,v_1,\dots,v_n)$ with $[s,v_i]=a_iv_i$. (It is unimodular iff $\sum a_i=0$.) Define $\lambda(s)=1$, $\lambda(v_i)=0$, so the weights are $0$ and the $a_i\lambda$ (which are strong weights). For $i\neq j$ we have $H_2(\mathfrak{g},V_{(a_i+a_j)\lambda})\neq 0$, although usually $(a_i+a_j)\lambda$ is not a weight; this gives unimodular counterexamples to the conjecture in dimension $n+1\ge 4$.

(e) Variant of (a): consider the Lie algebra with basis $(s,x,y,z)$ and nonzero brackets $[s,x]=ax$, $[s,y]=by$, $[s,z]=(a+b)z$, $[x,y]=z$, with $a,b,c$ scalars with $a,b,a+b\neq 0$. Let $\lambda$ map $s$ to $1$ and other basis elements to $0$. So the weights are the $t\lambda$ for $t\in\{0,a,b,a+b\}$.

Then a direct computation shows that the homology of $V_{t\lambda}$ is nonzero exactly for $t\in\{0,a,b,2a+b,a+2b,2a+2b\}$ (for $0$ the non-vanishing occurs in degree $0, 1$; for $a,b$ it occurs in degree $1$ and $2$; for $2a+b$ and $a+2b$ it occurs in degree $2$ and $3$ and for $2a+2b$ it occurs in degree $4$).

Thus in this example both implications of the conjecture fail: $(a+b)\lambda$ is a weight of the adjoint representation but $V_{(a+b)\lambda}$ has zero homology, while $t\lambda$ for $t\in\{2a+b,a+2b,2a+2b\}$ are not weights of the adjoint representation but for these values $V_{t\lambda}$ has homology.

  • $\begingroup$ Nice! But my question was on higher homologies too! (I'm sorry if that was unclear) I am interested when any(from 0 to nth) homologies vanish. $\endgroup$ – Boris Bilich May 7 at 5:08
  • $\begingroup$ OK for some reason I thought the focus was on $H_1$, so this is just partial information. Note that $H_0(\mathfrak{g},V_\lambda)=0$ iff $\lambda\neq 0$, so for $\lambda=0$ the homology does not vanish. $\endgroup$ – YCor May 7 at 6:49
  • $\begingroup$ Now in arbitrary degree I don't have a general description, but I now provide a 4-dimensional example (e) where both directions of the conjecture fail. $\endgroup$ – YCor May 7 at 12:51

To fill out my comment with a partial answer to the question.

Note that given any (left) $\mathfrak{g}$-module $V$ one can compute $H_i(\mathfrak{g},V)$ as $\mathrm{Tor}_i^{U(\mathfrak{g})}(\mathbb{C},V)$. If $S$ is any left Ore set that does not meet the augmentation ideal $I=\ker (U(\mathfrak{g})\to \mathbb{C})$ then the ring homomorphism $U(\mathfrak{g})\to \mathbb{C}$ factors through $U(\mathfrak{g})_S$. The functors $\mathbb{C}\otimes_{U(\mathfrak{g})} -$ and $\mathbb{C}\otimes_{U(\mathfrak{g})_S} (U(\mathfrak{g})_S \otimes_{U(\mathfrak{g})}-)$ are isomorphic and so have the same derived functors. Since $U(\mathfrak{g})_S$ is a flat right $U(\mathfrak{g})$-module (since $S$ was assumed a left Ore set) it follows that $$H_i(\mathfrak{g},V) = \mathrm{Tor}_i^{U(\mathfrak{g})_S}(\mathbb{C}, V_S)$$ where as usual $V_S$ denotes the localisation $U(\mathfrak{g})_S\otimes_{U(\mathfrak{g})}V$ of $V$ at $S$. In particular if $V_S=0$ then all homology groups of $V$ are trivial.

Now we come to the paper of Ken Brown numdam.org/article/CM_1984__53_3_347_0.pdf I mentioned in the comments. Theorem 5.3 tells us that the set Brown calls $\mathscr{S}'(I)$ is a left Ore set (here we use that $\mathfrak{g}$ is soluble). Chasing back through the paper we see that this set is the intersection $\mathscr{C}(I)\cap \bigcap_{\lambda\in L(I)} \mathscr{C}(\tau_\lambda(I))$. Here for a prime ideal $P$, $\mathscr{C}(P)$ denotes the elements of $U(\mathfrak{g})$ that are not zero divisors mod $P$, $L(I)$ is a sub-semigroup of $(\mathfrak{g}/[\mathfrak{g},\mathfrak{g}])^\ast$ generated by a certain explicit finite subset (containing at most $\dim \mathfrak{g}$ elements) of $\mathfrak{g}^{\ast}$ determined by the adjoint action of $\mathfrak{g}$ and $\tau_\lambda$ is the 'winding automorphism' of $U(\mathfrak{g})$ defined by $x\mapsto x+\lambda(x)$ for $x\in \mathfrak{g}$.

Now by Theorem 6.1 of the Brown paper if $S=\mathscr{S}'(I)$ then the localisation $U(\mathfrak{g})_S$ has countably many maximal one-sided ideals (which are all two sided) namely the ideals generated by the $\tau_\lambda(I)$ for $\lambda\in L(P)\cup\{0\}$. In follows that $(\mathbb{C}_\lambda)_S=0$ unless $\lambda\in L(P)\cup\{0\}$.

I believe that in the example in the question $L(P)$ consists of the elements of $(\mathfrak{g}/[\mathfrak{g},\mathfrak{g}])^\ast$ such that $x$ gets sent to a positive integer. So we see that this sufficient condition for vanishing homology is not necessary but gets you somewhere.

Perhaps I should close by noting that if $\mathfrak{g}$ is nilpotent then $L(P)=0$ and the fact that all homology groups $H_i(\mathfrak{g},V_\lambda)$ are trivial for any $\lambda\neq 0$ can be recovered.

  • $\begingroup$ It is possible that I've made sign errors near the end and that I mean that $(\mathbb{C}_{-\lambda})_S=0$ unless $\lambda\in L(P)\cup \{0\}$ and then $L(P)$ is the negation of what I claimed in the specific example. $\endgroup$ – Simon Wadsley May 7 at 20:25
  • $\begingroup$ If I were in the mood for making wild conjectures I might speculate that all homology groups of $\mathbb{C}_\lambda$ are trivial precisely if $\lambda$ is a sum of at most $\dim \mathfrak{g}/[\mathfrak{g,g}]$ of the generators of $L(I)$ given by Brown (if the sign was wrong before it is also wrong here). However I have no evidence for this. $\endgroup$ – Simon Wadsley May 7 at 20:38
  • $\begingroup$ I doubt this conjecture would match my example in (e) of my long post (for which the "bad" $\lambda$ are precisely $0$, $A$, $B$, $A+2B$, $2A+B$, and $2A+2B$ for suitable $A,B$, but not $A+B$). $\endgroup$ – YCor May 7 at 21:16
  • $\begingroup$ Agreed. In fact I think that when $P=I$, the augmentation ideal which is the important case here, the generators of $L(P)$ are precisely the weights of adjoint representation. I suppose it remains possible that the bad $\lambda$ are all a linear combination of at most $\dim \mathfrak{g}$ of these weights with repeats allowed (although the converse is not true). Perhaps even $H^i(\mathfrak{g},\mathbb{C}_\lambda)$ is zero unless $\lambda$ is a linear combination of $i$ of the weights (but not conversely). $\endgroup$ – Simon Wadsley May 8 at 8:24

This is an attempt to prove the (refined) conjecture I made in the comments of my previous answer. Let $\mathfrak{g}$ be a f.d soluble Lie algebra over $\mathbb{C}$. Let $\mathfrak{n}$ be its derived subalgebra which is nilpotent by Lie's theorem. I claim that for $\lambda\in ( \mathfrak{g}/\mathfrak{n})^\ast$, $H_i(\mathfrak{g}, \mathbb{C}_\lambda)\neq 0$ only if $\lambda$ is a sum of $i$ weights in the adjoint action.

To see this we pick a one-dimensional ideal $\mathfrak{z}=\mathbb{C}z$ in $ \mathfrak{g}$ contained in $\mathfrak{n}$. [If $\mathfrak{n}=0$ the result is easy by the now commutative methods in my previous answer.] Let's say z has weight $\mu$.

The Hochschild-Serre spectral sequence says that $H_i(\mathfrak{g}/\mathfrak{z} , H_j(\mathfrak{z} ,\mathbb{C}_\lambda))$ converges to $H_{i+j}(\mathfrak{g},\mathbb{C}_\lambda)$.

So we first understand $H_j(\mathfrak{z},\mathbb{C}_\lambda)$ as a $\mathfrak{g}$-module. This can be computed as $\mathrm{Tor}_j^{U(\mathfrak{g})}(U(\mathfrak{g})/(z), \mathbb{C}_\lambda)$ which is the $j$th homology group of $$0\to U(\mathfrak{g})\otimes_{U(\mathfrak{g})} \mathbb{C}_\lambda \to U(\mathfrak{g})\otimes_{U(\mathfrak{g})} \mathbb{C}_\lambda\to 0$$ where the non-zero map is multiplication by $z$. It follows that $$H_0(\mathfrak{z},\mathbb{C}_\lambda)=\mathbb{C}_\lambda$$ and $$H_1(\mathfrak{z}, \mathbb{C}_\lambda)=\mathbb{C}_{\lambda-\mu}.$$ By induction on $\dim\mathfrak{g}$ we're done.

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    $\begingroup$ Of course this also addresses Victor Protsak's question on another answer. $\endgroup$ – Simon Wadsley May 10 at 17:15

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