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Take the Verma module for the sl2 Lie algebra of highest weight $-2$, with basis $v_m$ and $f{v}_m=v_{m+1}$, $h{v}_m=(-2-2m)v_m$, $e{v}_m=-m(m+1)v_{m-1}$. It is simple. In ${\mathfrak{sl}}_2\otimes V$, the highest vectors of weight $-2$ are multiples of $w=e\otimes v_1-h\otimes v_0=f(e\otimes v_0)$. Every homomorphism to $V$ kills $e\otimes v_0$, whose weight is zero, and hence kills $w$. Every embedded copy of $V$ would have its highest vector proportional to $w$, so no projection onto such a copy can exist. Thus semisimplicity of the Lie algebra alone does not provide an action summand in infinite dimension.

Over a field of characteristic zero, a locally nilpotent operator has an algebraically defined exponential, because the sum is finite for each vector. Its inverse is $\exp (-T)$. The nilpotence bound may depend on the vector. In an sl2 Lie algebra Verma module, the raising operator is locally nilpotent but the lowering operator is not, so the latter's exponential does not define an endomorphism of the ordinary algebraic module.

// Target: lie-algebra.bigb

The Kostant partition function $P(\nu )$ counts the ways to express a weight $\nu$ as a nonnegative integer combination of positive roots. By the Poincaré-Birkhoff-Witt theorem, $P(\lambda -\mu )$ is the multiplicity of the weight $\mu$ in the Verma module $M(\lambda )$.

The Verma module $M(0)$ has basis $v_n=f^n{v}_0$, with $h{v}_n=-2n{v}_n$, $f{v}_n=v_{n+1}$ and $e{v}_n=n(1-n)v_{n-1}$. Its submodule spanned by $v_n$ for $n\ge 1$ is $M(-2)$, and the quotient is the trivial one-dimensional module. A complementary submodule would be trivial and hence lie in the zero-weight space $\mathbb{C}{v}_0$, but $f{v}_0=v_1\ne 0$. Therefore the displayed extension does not split. This explains why the finite-dimensional hypothesis in the Weyl complete reducibility theorem is essential.

Work first in a finite-dimensional SU(2) representation, with its invariant positive inner product. In the complexified Lie algebra, choose $H^{†}=H$ and $(E^{-}{)}^{†}=E^{+}$. Let $v_0$ be a highest-weight vector with $H{v}_0=m{v}_0$ and $E^{+}{v}_0=0$, and put $v_r=(E^{-}{)}^r{v}_0$. The commutator with $H$ givesInduction also givesFor $r=1$ this is $E^{+}{E}^{-}{v}_0=H{v}_0=m{v}_0$. For the induction step, use $E^{+}{E}^{-}=E^{-}{E}^{+}+H$ on $v_r$: the coefficient becomes $r(m-r+1)+m-2r=(r+1)(m-r)$. In particularwhich proves the assumed formula rather than merely invoking it.

The norm identity $\Vert E^{-}{v}_r{\Vert }^2=(r+1)(m-r)\Vert v_r{\Vert }^2$ shows that, for nonnegative integer $m$, the ladder is nonzero for $0\le r\le m$ and stops with $E^{-}{v}_m=0$. Distinct eigenvalues make these vectors independent. Their span is stable under $H,E^{\pm }$, and in the irreducible module it is the full space. ThusFor normalized states, $E^{-}|n⟩=\frac{1}{2}\sqrt{(m+n)(m-n+2)}|n-2⟩$, with the last state killed. General finite-dimensional SU(2) representations are direct sums of these ladders.

The representation category matters here. The Lie-algebra relations alone allow an infinite Verma module even at integral highest weight: its nonzero vector $v_{m+1}$ is singular. The finite representation can instead be constructed algebraically by taking the basis $v_0,\dots ,v_m$ with the displayed actions and declaring $E^{-}{v}_m=0$; all three Lie brackets then hold, including at the endpoints. Thus the finite-dimensional assertion uses the integrable SU(2) module, or this explicit finite quotient, rather than assuming that a singular vector must vanish in every highest-weight module.

For rank two, take a Cartan subalgebra $\mathfrak{h}$ with commuting basis elements. A root $\alpha$ is a nonzero linear functional on $\mathfrak{h}$ for which a root space contains an element $E_{\alpha }$ satisfyingChoose simple roots ${\alpha }_1,{\alpha }_2$ and their coroots $H_i={\alpha }_i^{\vee }$. In the PDF's index convention, the Cartan matrix isNormalize $E_i^{\pm }=E_{\pm {\alpha }_i}$ so $[E_i^{+},E_i^{-}]=H_i$. The difference of distinct simple roots is not a root, so the mixed positive-negative bracket is zero. This givesThe transpose in the last formula is essential and is printed correctly in the original PDF.

A highest-weight vector of labels $({\overline{n}}_1,{\overline{n}}_2)$ is killed by both $E_i^{+}$ and is an eigenvector of the two $H_i$. Repeated words in $E_1^{-},E_2^{-}$ generate its highest-weight representation; each lowers the weight by the corresponding simple root. Restricting to the sl2 subalgebra associated with a root for either simple root gives the preceding integer ladder, so finite-dimensional highest labels must be nonnegative integers. The finite irreducible module is the highest-weight quotient imposing $(E_i^{-}{)}^{{\overline{n}}_i+1}v=0$ and the submodules generated by these relations, together with the Serre relations. Choosing independent nonzero lowering words gives a simultaneous weight basis; an extra multiplicity index is needed where a weight space has dimension greater than one.

For the specified B2 root system, the simple-root squared lengths are $1,2$ and their inner product is $-1$, givingThe positive root ${\alpha }_1+{\alpha }_2$ is generated by $[E_1^{+},E_2^{+}]\ne 0$, and $2{\alpha }_1+{\alpha }_2$ by $[E_1^{+},[E_1^{+},E_2^{+}]]\ne 0$. No further positive sums occur, so $(\mathrm{ad}{E}_1^{+}{)}^3{E}_2^{+}=0$ and $(\mathrm{ad}{E}_2^{+}{)}^2{E}_1^{+}=0$. These are the relevant Serre relations. Applying the Cartan commutators to a weight vector proveswhenever the image is nonzero.

To enumerate the requested representation without guessing when arrows stop, realize it concretely as the Adjoint representation. For a Euclidean weight $(x,y)$, its labels are $(n_1,n_2)=(2x,-x+y)$. The highest root $2{\alpha }_1+{\alpha }_2=(1,1)$ has labels $(2,0)$, so it generates the irreducible adjoint module of this simple rank-two Lie algebra. Its nonzero weights are exactly its eight roots, and its zero weight space is the two-dimensional Cartan subalgebra. This gives the B2 adjoint representation weight diagram and the nine distinct states labels:Their actual production by lowering can be seen fromAn arrow labelled $i$ denotes a nonzero action of $E_i^{-}$ between the indicated weight spaces. The two paths into zero produce independent Cartan vectors. Hence there are nine distinct pairs, as requested when degeneracy is disregarded, but the full representation has dimension ten, not nine.

Let $v_0$ be the highest vector of the Verma module $M(n)$ and set $v_k=f^k{v}_0$. The Poincaré-Birkhoff-Witt theorem implies that $(v_k{)}_{k\ge 0}$ is a basis, so none of these vectors vanishes. The actions areThe raising operator is a locally nilpotent operator: on every vector, repeated application of $e$ eventually removes all its finitely many basis terms. Thus its algebraic exponential of a locally nilpotent operator is defined, and $\exp (e)v_0=v_0$. But the middle factor in the proposed product would have to send this vector towhich has infinitely many nonzero, linearly independent basis coefficients. An element of the algebraic Verma module is a finite linear combination of the $v_k$, so this expression is not in $M(n)$. In particular the lowering operator is not locally nilpotent.

Even for integral $n\ge 0$, $f^{n+1}{v}_0$ is nonzero in the Verma module; it becomes zero only in its finite-dimensional irreducible quotient $L(n)$. Hence $s$ is not defined as the displayed composition of endomorphisms of the algebraic Verma module. Working in a completion would be a different problem and is not part of the module specified here.

Give $\mathfrak{g}$ its Adjoint representation and use the tensor product of Lie algebra representations. Define the action map of a Lie algebra representationFor $y\in \mathfrak{g}$, the action on the source is $y(x\otimes v)=[y,x]\otimes v+x\otimes yv$. The defining representation identity then givesThus $a$ is a Lie algebra representation homomorphism.

If $V$ is nontrivial and simple, the action is not identically zero, so the image of $a$ is a nonzero submodule. Simplicity forces $a$ to be surjective. For a finite-dimensional $V$ and a semisimple Lie algebra, the Weyl complete reducibility theorem provides an invariant complement $U$ to $\ker a$ in $\mathfrak{g}\otimes V$. The restriction $a{|}_U$ is then an isomorphism onto $V$, givingThis proves the requested summand assertion in the finite-dimensional setting.

If finite dimensionality is not implicit, the final assertion needs that hypothesis: the action summand can fail for an infinite-dimensional simple module. Here is an explicit counterexample. Take $\mathfrak{g}={\mathfrak{sl}}_2$ with $[h,e]=2e$, $[h,f]=-2f$, $[e,f]=h$, and the Verma module of highest weight $-2$:The lowering coefficient never vanishes for $m>0$, so an invariant subspace, using the distinct $h$-weights to isolate a basis vector and then raising, contains $v_0$ and hence all of $V$. Thus $V$ is simple.

In ${\mathfrak{sl}}_2\otimes V$, a vector of weight $-2$ is $Ae\otimes v_1+Bh\otimes v_0$. Applying $e$ gives $-2(A+B)e\otimes v_0$, so its highest vectors are precisely the multiples of $w=e\otimes v_1-h\otimes v_0$. But $w=f(e\otimes v_0)$, and $e\otimes v_0$ has weight zero, absent from $V$. Every homomorphism $r:\mathfrak{g}\otimes V\to V$ therefore kills this vector and then kills $w$. Any embedded copy of $V$ would send its highest vector to a nonzero multiple of $w$, so it cannot admit a projection back to $V$. This rules out a direct summand, not merely a splitting of the particular action map.

Choose a base of a root system and its positive roots, so that the triangular decomposition of a Lie algebra is $L={\mathfrak{n}}^{-}\oplus H\oplus {\mathfrak{n}}^{+}$. A primitive element of a Lie algebra representation of weight $\omega \in H^{\ast }$ is a nonzero vector $v$ such thatEquivalently it is a highest-weight vector. Being primitive depends on the chosen positive system; no integrality condition is part of this definition.

Construct the Verma module for an arbitrary $\omega$. The subalgebra $\mathfrak{b}=H\oplus {\mathfrak{n}}^{+}$ has a one-dimensional module ${\mathbb{C}}_{\omega }$ on which $H$ acts by $\omega$ and ${\mathfrak{n}}^{+}$ acts by zero. This is a representation because $[\mathfrak{b},\mathfrak{b}]\subseteq {\mathfrak{n}}^{+}$. Setwhere $U$ denotes the universal enveloping algebra. The Poincaré-Birkhoff-Witt theorem, applied with negative-root vectors first, then Cartan vectors and positive-root vectors, givesas vector spaces. In particular $v_{\omega }=1\otimes 1$ is nonzero, generates $M(\omega )$, and is primitive of weight $\omega$.

The same ordered monomials show that $M(\omega )$ is a direct sum of weight spaces, with weights of the formand that its weight-$\omega$ space is exactly $\mathbb{C}{v}_{\omega }$. Each vector has only finitely many weight components. If a submodule contains a vector with nonzero weight-$\omega$ component, choose $h\in H$ that separates $\omega$ from the finitely many other weights of that vector. Polynomial interpolation in the action of $h$ extracts a nonzero multiple of $v_{\omega }$. Such a submodule is all of $M(\omega )$, because $v_{\omega }$ generates it. Consequently every proper submodule has zero component in weight $\omega$.

Let $N$ be the sum of all proper submodules. Every finite sum still has zero weight-$\omega$ component, so $N$ is proper; it contains every proper submodule and is therefore the unique maximal proper submodule. Its quotientis irreducible, and the nonzero image of $v_{\omega }$ is primitive of weight $\omega$. This proves existence for every $\omega \in H^{\ast }$ and gives the irreducible quotient of a Verma module. It also proves uniqueness up to isomorphism among irreducible modules generated by a primitive vector of that weight: the defining relations induce a surjection from $M(\omega )$, whose proper kernel must be $N$.

These modules are not generally finite-dimensional. For example, in an ${\mathfrak{sl}}_2$ root subalgebra, a primitive vector with $hv=\lambda v$ satisfiesIf its irreducible highest-weight module is finite-dimensional, the weight-lowering sequence eventually stops. For the first $m$ with $f^{m+1}v=0$ and $f^{m}v\ne 0$, the identity forces $\lambda =m\in {\mathbb{Z}}_{\ge 0}$. In general a finite-dimensional highest-weight module therefore requires nonnegative integral values on all simple coroots; arbitrary $\omega$ in the request must allow infinite-dimensional representations. This is consistent with the finite-dimensional qualification made in Question 2.

For simple coroots $h_1,\dots ,h_r$ forming a basis of $H$, the fundamental weights are the dual basis ${\omega }_1,\dots ,{\omega }_r\in H^{\ast }$:For the special linear Lie algebra ${\mathfrak{sl}}_3(\mathbb{C})$, take the diagonal trace-zero Cartan subalgebra and positive roots corresponding to upper triangular matrix units. Write ${\epsilon }_i(\mathrm{diag}(t_1,t_2,t_3))=t_i$. The simple roots are ${\alpha }_1={\epsilon }_1-{\epsilon }_2$, ${\alpha }_2={\epsilon }_2-{\epsilon }_3$, and the simple coroots areThe fundamental weights of sl3 are thereforeChecking these on $h_1,h_2$ gives the two coordinate vectors $(1,0)$ and $(0,1)$. In the Euclidean trace-zero plane they can also be writtenThe Fundamental representations of sl3 make these weights concrete: ${\mathbb{C}}^3$ has primitive vector $e_1$ of weight ${\omega }_1$, and ${\Lambda }^2{\mathbb{C}}^3$ has primitive vector $e_1\wedge e_2$ of weight ${\omega }_2$. Every finite-dimensional irreducible highest weight is $a{\omega }_1+b{\omega }_2$ with nonnegative integers $a,b$.

Choose a positive system of a root system and the corresponding triangular decomposition of a Lie algebra $L={\mathfrak{n}}^{-}\oplus H\oplus {\mathfrak{n}}^{+}$. A primitive element of a Lie algebra representation of weight $\omega \in H^{\ast }$ is a nonzero vector $v$ with $hv=\omega (h)v$ for every $h\in H$ and ${\mathfrak{n}}^{+}v=0$. Thus it is a highest-weight vector; the choice of positive roots is part of this definition. A highest-weight representation is generated by such a vector.

Let $\mathfrak{b}=H\oplus {\mathfrak{n}}^{+}$ be the corresponding Borel subalgebra. Define its one-dimensional Lie algebra representation ${\mathbb{C}}_{\omega }$ by the scalar $\omega (h)$ on $H$ and zero on ${\mathfrak{n}}^{+}$. This respects the Lie bracket, since $[\mathfrak{b},\mathfrak{b}]\subseteq {\mathfrak{n}}^{+}$. The induced Verma moduleis nonzero: the Poincaré-Birkhoff-Witt theorem identifies its underlying vector space with $U({\mathfrak{n}}^{-})$, and $v=1\otimes 1$ is a primitive element of weight $\omega$. Its weight spaces are finite dimensional, its top weight space is the line $\mathbb{C}v$, and all other weights are $\omega -\sum n_i{\alpha }_i$ with $n_i\ge 0$ and at least one positive coefficient.

A proper submodule cannot contain $v$, because $v$ generates $M(\omega )$. More strongly it has no component of weight $\omega$: any finite sum of distinct weight vectors can be projected onto its individual components by a polynomial in a generic element of $H$. Therefore the sum $N$ of all proper submodules still misses the top weight line and is proper. It contains every proper submodule, so the irreducible quotient of a Verma moduleis irreducible and retains the nonzero primitive element $v+N$. This constructs the requested representation for every $\omega \in H^{\ast }$. It is not asserted to be finite dimensional for arbitrary $\omega$.

For the sl2 Lie algebra, use $h=\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)$, $e=\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)$ and $f=\left(\begin{array}{cc}0& 0\\ 1& 0\end{array}\right)$, with $[h,e]=2e$, $[h,f]=-2f$, $[e,f]=h$. The classification of finite-dimensional sl2 representations gives one irreducible $V_n$ for every integer $n\ge 0$. In a basis $v_0,\dots ,v_n$ its action iswhere $v_{-1}=v_{n+1}=0$. These formulas satisfy the three Lie brackets. Any nonzero invariant subspace contains a weight vector by polynomial projection using $h$; repeated application of $e$ gives $v_0$, and applications of $f$ then give the entire basis. Thus $V_n$ is irreducible.

Conversely, in any finite-dimensional irreducible module, start with an eigenvector of $h$ and apply $e$ until reaching a nonzero vector $v$ with $ev=0$. This process terminates because $e$ increases the eigenvalue by two and only finitely many eigenvalues occur. Write $hv=av$. The sl2 highest-weight lowering formula givesLet $f^{n}v\ne 0$ and $f^{n+1}v=0$, which again follows from the finite set of weights. Applying $e$ to the latter identity yields $(n+1)(a-n)f^{n}v=0$, so $a=n$. The resulting $n+1$ distinct weight vectors span an invariant subspace, hence the entire irreducible module. Therefore $\dim V_n=n+1$ and its primitive weight is ${\omega }_n(h)=n$, with $n=0,1,2,\dots$; its weights are $n,n-2,\dots ,-n$.

The Weyl complete reducibility theorem states that every finite-dimensional Lie algebra representation of a finite-dimensional complex semisimple Lie algebra is a direct sum of Irreducible Lie algebra representations. Equivalently, every invariant vector subspace has an invariant complement.

We use the permitted Casimir operator properties in the following precise form. There is a central quadratic operator $C$ commuting with the action on every module; it is zero on the trivial Lie algebra representation, and on every nontrivial finite-dimensional Irreducible Lie algebra representation it is a nonzero scalar $c$. This follows from the Schur lemma and the Casimir eigenvalue $c_{\lambda }=(\lambda ,\lambda +2\rho )$, using the Killing form normalization. We also use the permitted one-dimensional-representation fact: a complex semisimple Lie algebra has only trivial one-dimensional representations. Equivalently, it is a perfect Lie algebra, $[\mathfrak{g},\mathfrak{g}]=\mathfrak{g}$, so a character $\mathfrak{g}\to \mathbb{C}$ annihilating brackets must vanish. Neither fact assumes complete reducibility of the module being proved reducible.

First prove that every finite-dimensional short exact sequencewith trivial quotient splits. If $W$ is nontrivial irreducible, the Casimir operator $C_E$ has image in $W$ and restricts to $c{I}_W$ there. Consequently $\ker C_E$ is a one-dimensional invariant complement to $W$. If $W$ is trivial irreducible, $E$ has a basis in which every action is $\left(\begin{array}{cc}0& a(x)\\ 0& 0\end{array}\right)$. The Lie algebra representation identity makes $a([x,y])=0$, so the one-dimensional-representation fact gives $a=0$ and again the sequence splits.

For general $W$, induct on $\dim W$. Choose an irreducible submodule $W_1\subseteq W$. The induced sequence with kernel $W/W_1$ and middle term $E/W_1$ splits by induction. The inverse image $E_1\subseteq E$ of its invariant complement is a submodule fitting into $0\to W_1\to E_1\to \mathbb{C}\to 0$. The irreducible-kernel case gives an invariant line in $E_1$ mapping isomorphically to the quotient. It is also an invariant complement to $W$ in $E$. The case $W=0$ starts this induction. This proves splitting of a trivial quotient for a semisimple Lie algebra, including kernels that are not assumed completely reducible.

Now let $U\subseteq V$ be any invariant subspace. On the Hom representation $\mathrm{Hom}(V,U)$ the action isLet $E$ consist of the maps whose restriction to $U$ is a scalar multiple of $I_U$. This is a submodule, and restriction givesThe right-hand map is surjective because an ordinary linear projection $V\to U$ exists; its quotient action is trivial because a commutator with $I_U$ is zero. The splitting just proved supplies an invariant $T\in E$ with $T{|}_U=I_U$. Thus $T$ intertwines the actions, $T^2=T$, andThe cases $U=0$ and $U=V$ are immediate. Choosing an irreducible submodule and repeating this complement construction proves the Weyl complete reducibility theorem. This last step is invariant complement from an equivariant projection.

Dropping finite dimensionality gives an example with the complex simple Lie algebra ${\mathfrak{sl}}_2$. Its Verma module of highest weight zero has a basis $v_0,v_1,\dots$ withwhere $e{v}_0=0$. These actions obey $[h,e]=2e$, $[h,f]=-2f$, $[e,f]=h$. The span of $v_1,v_2,\dots$ is a proper submodule and the quotient is trivial. It has no invariant complement: such a complement would be a trivial line, while $f$ is injective on the entire module. Thus a simple Lie algebra can have an infinite-dimensional representation that is not completely reducible, even over $\mathbb{C}$.

The Poincaré-Birkhoff-Witt theorem shows that the weights of the Verma module $M({\omega }_2)$ arewith multiplicities given by the corresponding Kostant partition function.

The Dynkin labels of ${\omega }_2$ are $(0,1)$. Hence the two simple-root singular vectors are $f_1{v}_{{\omega }_2}$, of weight ${\omega }_2-{\alpha }_1$, and $f_2^2{v}_{{\omega }_2}$, of weight ${\omega }_2-2{\alpha }_2$. They generate the Maximal proper submodule of a dominant integral Verma module. Its set of weights is consequently

Choose a Borel subalgebra $\mathfrak{b}=\mathfrak{t}\oplus {\mathfrak{n}}^{+}$. For $\lambda \in {\mathfrak{t}}^{\ast }$, let ${\mathbb{C}}_{\lambda }$ be the one-dimensional $\mathfrak{b}$-module on which ${\mathfrak{n}}^{+}$ acts by zero and $h\in \mathfrak{t}$ acts by $\lambda (h)$. The Verma module isIts universal property of a Verma module says that any vector of weight $\lambda$ annihilated by ${\mathfrak{n}}^{+}$ receives the canonical highest-weight vector under one unique module homomorphism from $M(\lambda )$.

The sum of the proper submodules of $M(\lambda )$ is its unique maximal proper submodule, because no proper submodule contains the highest-weight vector. Its quotient $V(\lambda )$ is therefore the unique irreducible quotient of a Verma module, and hence the unique irreducible highest-weight module of weight $\lambda$.

The module $V(\lambda )$ is finite-dimensional exactly when $\lambda$ is a dominant integral weight:for every simple root ${\alpha }_i$.

Choose the Borel subalgebra $\mathfrak{b}=\mathfrak{t}\oplus {\mathfrak{n}}^{+}$ determined by the positive roots. Regard the one-dimensional space ${\mathbb{C}}_{\lambda }$ as a $\mathfrak{b}$-module on which ${\mathfrak{n}}^{+}$ acts by zero and $h\in \mathfrak{t}$ acts by $\lambda (h)$. The Verma module isThe Poincaré-Birkhoff-Witt theorem identifies it as a vector space with $U({\mathfrak{n}}^{-})$ acting on a highest-weight vector $v_{\lambda }$.

For each positive root $\alpha$, arbitrary powers of a negative-root vector contribute the geometric series $1+e^{-\alpha }+e^{-2\alpha }+\cdots$. Consequently the formal character of a weight module isThis product is interpreted in the completion of the group algebra in the negative-root direction; the PBW basis proves that every coefficient is the correct finite weight space dimension.

Choosing a positive system of a root system for a complex semisimple Lie algebra divides its root spaces into negative and positive ones. Their respective sums ${\mathfrak{n}}^{-}$ and ${\mathfrak{n}}^{+}$, together with the Cartan subalgebra $\mathfrak{h}$, give this direct sum of vector spaces. Both ${\mathfrak{n}}^{\pm }$ are nilpotent Lie algebras, and $\mathfrak{h}\oplus {\mathfrak{n}}^{+}$ is a Borel subalgebra. The Poincaré-Birkhoff-Witt theorem turns this into an ordered vector-space decomposition of the universal enveloping algebra, useful for building Verma modules.

An irreducible unitary highest weight module of the Virasoro algebra with $c<1$ has integer $m\ge 2$, $1\le r\le m-1$, $1\le s\le m$, and the displayed weights. Labels $(r,s)$ and $(m-r,m+1-s)$ represent the same module. It is obtained from the Verma module by quotienting its zero-norm submodules; the two generating singular-vector levels are $rs$ and $(m-r)(m+1-s)$. For $m=2$ this is the trivial $c=h=0$ module. For $m=3$, the three weights are $0$, $1/2$, $1/16$.