Recall that the Virasoro algebra can be decomposed as We can define a “universal” highest weight representation on it, called the Verma module.
Verma Module
Let , and fix complex numbers (the central charge) and (the highest weight).
Define a -module, whose underlying vector space is , with action
The Verma module, denoted or , is then the -module defined by the tensor productequipped with the left action
In other words, the Verma module is the initial object in the category of pointed highest weight representations (i.e., objects are tuples where is a highest weight vector) of weight .
Suppose is a complex vector space. We call a highest (lowest) weight representation for of weight for complex numbers if the following conditions are satisfied:
diagonalizes on with eigenvalues contained in ;
Eigenspaces of are finite-dimensional, in particular, the -eigenspace is one-dimensional;
Lemma
For any representation of a complex Lie algebra, with elements satisfying . Let . Then for all .
Proof For any , we have so .
Proposition
The Verma module is a highest weight representation for of weight . It has a unique nonzero proper irreducible quotient .
Proof By the Poincaré–Birkhoff–Witt theorem, is spanned by ordered monomials. So we can express any as where , and are basis of , and respectively. If is nonempty (in terms of words), then Therefore, it suffices to consider only the case when is empty, i.e., .
By our construction of the Verma module, is an eigenvector of generating the -eigenspace. Let us denote . Note that , so by the above lemma, It follows that any is an eigenvector of with eigenvalue for some . Hence, diagonalizes on . Moreover, there are only finite ways to decompose , so the dimension of each eigenspace is finite.
We now show that it has a unique nonzero proper irreducible quotient . In , the eigenspace is one-dimensional. Hence every proper submodule cannot intersect with . Note that the sum of any proper submodule is still proper, so it is possible to pick a unique maximal module by taking the sum of all proper submodules. Then is the unique nonzero proper irreducible quotient.
The Shapovalov Form
Lemma
Let be the weight of such that
Then , where is the left ideal .
Proof Define by , which is clearly surjective. For any , therefore , so . Conversely, describes the only relation we imposed when constructing (this is the universal property of ), concluding that , so by the first isomorphism theorem, .
Proposition
The unique quotient of admits a unique nonzero contravariant form up to scalar multiplication.
Proof Let us first construct a contravariant form on . By the Poincaré–Birkhoff–Witt theorem, any element of has either no factor from , or at least one factor from appearing on the left, so we can write . Denote and as the projections. We extend the anti-involution to by , and .
We claim that is -equivariant and multiplicative. The anti-involution fixes , and swaps and , so for all . Using the triangular decomposition of , we can write any as . Then so is multiplicative.
Now we can define on as It is anti-linear: . Moreover, , and it is contravariant by construction, so it is indeed a contravariant Hermitian form on .
The above lemma shows that can be pushed down to if it kills . That is, for all and , we want to show that Observe that proving the first equality. The second equality holds similarly. Hence we can define a contravariant form on by Now we show that is unique up to a scalar. Let be a highest weight vector. Then for any , a contravariant form must satisfy which is completely determined by .
Hence, we are able to define a distinguished contravariant form on :
Shapovalov Form
The Shapovalov form on the Verma module is the unique contravariant Hermitian form such that for .