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Given a (finite-dimensional) Lie algebra $\mathfrak{g}$, a subalgebra $\mathfrak{h}\subset\mathfrak{g}$, and a representation $\rho:\mathfrak{h}\rightarrow\mathfrak{gl}(V)$ of $\mathfrak{h}$, one can form (so I'm told) a representation of the whole algebra $\mathfrak{g}$ on

$$ \text{Ind}^{\mathfrak{g}}_{\mathfrak{h}}(V):=U(\mathfrak{g})\otimes_{U(\mathfrak{h})} V, $$

where $U(-)$ denotes the universal enveloping algebra of $-$. My question is simple:

How, exactly, is the representation of $\mathfrak{g}$ on $\text{Ind}^{\mathfrak{g}}_{\mathfrak{h}}(V)$ defined? Are there any good references covering this topic (focusing on the Lie algebra side, rather than induced representations of Lie groups, as essentially all references I've found discuss)?

In particular, references going through numerous examples actually computing the universal enveloping algebras and the induced representation would be tremendously appreciated.

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    $\begingroup$ It comes from $\mathfrak{g}$ acting on $U(\mathfrak{g})$ by left multiplication. A common special case of this is the construction of Verma modules; look that up for more details. $\endgroup$ – Qiaochu Yuan Jan 3 '16 at 5:52
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I assume that we are dealing with Lie algebras over an field $K$. Let me try to give an overview about how this induced representation is constructed, which is an exercise in understanding how representations of the Lie algebra $\mathfrak{g}$ relate to modules of the universal enveloping algebra $\mathcal{U}(\mathfrak{g})$. A book which you might want to look into is James E. Humphreys’ Introduction to Lie Algebras and Representation Theory, where the topic is covered from an algebraic point of view and without the use of Lie groups.

The universal enveloping algebra

As you probably already know the universal eneveloping algebra $\mathcal{U}(\mathfrak{g})$ of the Lie algebra $\mathfrak{g}$ is an associative, untial $K$-algebra, which can be defined as the quotient algebra $T(\mathfrak{g})/I$ where $T(\mathfrak{g})$ is the tensor algebra over $\mathfrak{g}$ and $I$ the two-sided ideal $I = \langle x \otimes y - y \otimes x - [x,y] \mid x,y \in \mathfrak{g}\rangle$. So as a vector space $\mathcal{U}(\mathfrak{g})$ is generated by the monomials $x_1 \dotsm x_n$ with $x_1, \dotsc, x_n \in \mathfrak{g}$. The inclusion $\mathfrak{g} \hookrightarrow \mathcal{U}(\mathfrak{g})$ is a homomorphism of Lie algebras, so we can regard $\mathfrak{g}$ as a Lie subalgebra of $\mathcal{U}(\mathfrak{g})$. The multiplication in $\mathcal{U}(\mathfrak{g})$ satisfies the property that $xy-yx = [x,y]$ for all $x,y \in \mathfrak{g} \subseteq \mathcal{U}(\mathfrak{g})$ (where the right hand side denotes the Lie bracket of $\mathfrak{g}$).

An essential property that follows from this construction of $\mathcal{U}(\mathfrak{g})$ is the Poincaré–Birkhoff–Witt theorem, which decribes a basis of $\mathcal{U}(\mathfrak{g})$.

Theorem (Poincaré–Birkhoff–Witt): Let $\mathfrak{g}$ be a Lie algebra and $(x_i)_{i \in I}$ a basis of $\mathfrak{g}$ where $(I, \leq)$ is a totally ordered set. Then the ordered monomials $$ x_{i_1} \dotsm x_{i_n} \qquad i_1, \dotsc, i_n \in I,\; i_1 \leq \dotsb \leq i_n $$ form a basis of $\mathcal{U}(\mathfrak{g})$.

Remark: This basis can also be written as $$ x_{i_1}^{p_1} \dotsm x_{i_n}^{p_n} \qquad i_1, \dotsc, i_n \in I, \; i_1 < \dotsb < i_n, \; p_1, \dotsc, p_n \in \mathbb{N}. $$

Example: The Lie algebra $\mathfrak{sl}_2(\mathbb{C})$ has a basis $(e,h,f)$ given by $$ e = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}, \; h = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}, \; f = \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}. $$ This basis is already totally ordered by the order in which we write the tupel $(e,h,f)$. By the PBW theorem it follows that $\mathcal{U}(\mathfrak{sl}_2(\mathbb{C}))$ has a basis consisting of the the monomials $e^l h^m f^n$ with $l,m,n \in \mathbb{N}$.

The nice thing about the universal enveloping algebra is its universal property:

Let $A$ be an associative and unital $K$-algebra. Then any Lie algebra homomorphism $\phi \colon \mathfrak{g} \to A$ extends uniquely to an algebra homomorphism $\Phi \colon \mathcal{U}(\mathfrak{g}) \to A$.

Representations of $\mathfrak{g}$ and $\mathcal{U}(\mathfrak{g})$-modules

This universal property has nice effects for the representation theory of $\mathfrak{g}$. A representation of $\mathfrak{g}$ is, by definition, a Lie algebra homomorphism $\mathfrak{g} \to \mathfrak{gl}(V)$ for some vector space $V$. Using the universal property of $\mathcal{U}(\mathfrak{g})$ it follows that the Lie algebra homomorphisms $\mathfrak{g} \to \mathfrak{gl}(V)$ correspond one-to-one to the algebra homomorphisms $\mathcal{U}(\mathfrak{g}) \to \mathrm{End}_K(V)$ via restricting and extending the homomorphisms in question.

More explicitely: If $\rho \colon \mathfrak{g} \to \mathfrak{gl}(V)$ is a representation of $\mathfrak{g}$ then for every $x \in \mathfrak{g}$ we have a multiplication (i.e. a bilinear map) $\mathfrak{g} \times V \to V$, $(x,v) \mapsto x.v$ via $x.v = \rho(x)(v)$ for all $x \in \mathfrak{g}$ and $v \in V$. This multiplication now extends to $\mathcal{U}(\mathfrak{g})$ as $$ (x_1 \dotsm x_n).v = x_1.(x_2.(\dotsm(x_{n-1}.(x_n.v)))) \quad \text{for all $x_1, \dotsc, x_n \in \mathfrak{g}$, $v \in V$}. $$ This results in a unique bilinear map $\mathcal{U}(\mathfrak{g}) \times V \to V$, $(x,v) \mapsto x.v$. With this $V$ is an $\mathcal{U}(\mathfrak{g})$-module. Notice that for $\mathfrak{g} \subseteq \mathcal{U}(\mathfrak{g})$ this module structure coincides with the representation we started with.

Extension of scalars

We will need the tensor product, and in particular its use for the so called extension of scalars. A detailed explanation can, for example, be found in Abstract Algebra by Dummit and Foote (see here for the section) .

The basic idea behind the induced representation is the extension of scalars: Suppose $R$ is a unitial ring, $S \subseteq R$ a subring (also unital) and $M$ a (unital) $S$-module. Then we want to somehow extend the $S$-module structure of $M$ to an $R$-module structure on $M$. The problem is that there is a priori no good way to extends the multiplication $S \times M \to M$ to a multiplication $R \times M \to M$. So instead, we replace $M$ by another abelian group, namely $R \otimes_S M$.

Recall that $R \otimes_S M$ is an abelian group generated by elements of the form $r \otimes m$ with $r \in R$ and $m \in M$, under the constraints that $$ (r+r') \otimes m = r \otimes m + r' \otimes m, \\ r \otimes (m+m') = r \otimes m + r \otimes m', \\ (rs) \otimes m = r \otimes (sm) $$ where $r, r' \in R$, $m, m' \in M$ and $s \in S$.

The nice thing about $R \otimes_S M$ is that it naturally carries the structure of an $R$-module via $$ r' . (r \otimes m) = (r' r) \otimes m $$ for $r', r \in R$ and $m \in M$. Notice that for all $s \in S$ and $m \in M$ we have $$ s . (1 \otimes m) = (s \cdot 1) \otimes m = s \otimes m = (1 \cdot s) \otimes m = 1 \otimes (sm), $$ so we can, to some extend, thing of $R \otimes_S M$ as an extension of $M$. Indeed, the homomorphism $M \to R \otimes_S M$, $m \mapsto 1 \otimes m$ is a homomorphism of $S$-modules. (A word of warning though: It is not always true that this homomorphism is injective. So we can not necessarily regard $M$ as an $S$-submodule of $R \otimes_S M$.)

We refer to this process of “extending” the $S$-module structure of $M$ to the $R$-module structure on $R \otimes_S M$ as extension of scalars.

The induced representation

We are now ready to tackle the induced representation. For this let $\mathfrak{g}$ be a Lie algebra, $\mathfrak{h} \subseteq \mathfrak{g}$ a Lie subalgebra and $\rho \colon \mathfrak{h} \to \mathfrak{gl}(V)$ a representation of $\mathfrak{h}$. Then for all $x \in \mathfrak{h}$ and $v \in V$ we have the product $x.v = \rho(x)(v)$.

Notice that the inclusion $\mathfrak{h} \hookrightarrow \mathfrak{g}$, which is a homomorphism of Lie algebras, induces an algebra homomorphism $\mathcal{U}(\mathfrak{h}) \to \mathcal{U}(\mathfrak{g})$ with $$ x_1 \dotsm x_n \mapsto x_1 \dotsm x_n \quad \text{for all $x_1, \dotsc, x_n \in \mathfrak{h}$}. $$ It is easy to see that this algebra homomorphism is injective, for example by using the PBW theorem. So we can regard $\mathcal{U}(\mathfrak{h})$ as a subalgebra of $\mathcal{U}(\mathfrak{g})$.

On an abstract level we can now simply apply extension of scalars: That $V$ is a representation of $\mathfrak{h}$ means that it is a $\mathcal{U}(\mathfrak{h})$-module via $$ (x_1 \dotsm x_n).v = x_1.(x_2.(\dotsm(x_{n-1}.(x_n.v)))) \quad \text{for all $x_1, \dotsc, x_n \in \mathfrak{h}$, $v \in V$}. $$ So by applying extension of scalars we get the $\mathcal{U}(\mathfrak{g})$-module $\mathrm{Ind}_\mathfrak{h}^\mathfrak{g}(V) = \mathcal{U}(\mathfrak{g}) \otimes_{\mathcal{U}(\mathfrak{h})} V$.

But let’s see how the $\mathcal{U}(\mathfrak{g})$-module structure on $\mathrm{Ind}_\mathfrak{h}^\mathfrak{g}(V)$ really looks like: As a vector space $\mathrm{Ind}_\mathfrak{h}^\mathfrak{g}(V)$ is generated by the simple tensors $x \otimes v$ with $x \in \mathcal{U}(\mathfrak{g})$ and $v \in V$. Because $\mathcal{U}(\mathfrak{g})$ is generated by the monomial $x_1 \dotsm x_n$ with $x_1, \dotsc, x_n \in \mathfrak{g}$ it follows that $\mathrm{Ind}_\mathfrak{h}^\mathfrak{g}(V)$ is, as a vector space, generated by the elements $$ (x_1 \dotsm x_n) \otimes v \quad \text{with $x_1, \dotsc, x_n \in \mathfrak{g}$, $v \in V$}. $$ In terms of these generators the $\mathcal{U}(\mathfrak{g})$-module structure of $\mathrm{Ind}_\mathfrak{h}^\mathfrak{g}(V)$ is given by the multiplication $$ (x_1 \dotsm x_n) . ((y_1 \dotsm y_m) \otimes v) = (x_1 \dotsm x_n \cdot y_1 \dotsm y_m) \otimes v. $$ The fact that we are tensoring over $\mathcal{U}(\mathfrak{h})$ has the effect that $$ h . (1 \otimes v) = h \otimes v = 1 \otimes (h.v) \quad \text{for all $h \in \mathfrak{h}$, $v \in V$}. $$

This properties can nicely be put together by choosing a fitting basis of $\mathfrak{g}$ and using the PBW theorem: Let $(b_i)_{i \in I}$ be a basis of $\mathfrak{g}$ such that $(I,\leq)$ is a totally ordered set, and such that we have a partition $I = J' \cup J$ so that $(b_j)_{j \in J}$ is a basis of $\mathfrak{h}$ and $i \leq j$ for all $i \in J'$ and $j \in J$.

Then by the PBW-theorem the ordered monmials $$ b_{i_1} \dotsm b_{i_n} b_{j_1} \dotsm b_{j_m} \qquad \begin{gathered} i_1, \dotsc, i_n \in J', \, i_1 \leq \dotsb \leq i_n, \\ j_1, \dotsc, j_m \in J, \, j_1 \leq \dotsb \leq _m \end{gathered} $$ form a basis of $\mathcal{U}(\mathfrak{g})$, and the ordered monomials $$ b_{j_1} \dotsm b_{j_m} \qquad j_1, \dotsc, j_m \in J, \, j_1 \leq \dotsb \leq j_m $$ form a basis of the subalgebra $\mathcal{U}(\mathfrak{h})$. With respect to this basis the $\mathcal{U}(\mathfrak{g})$-module structure on $\mathrm{Ind}_\mathfrak{h}^\mathfrak{g}(V)$ is given by $$ (b_{i_1} \dotsm b_{i_n} b_{j_1} \dotsm b_{j_m}) . (1 \otimes v) = (b_{i_1} \dotsm b_{i_n}) \otimes (b_{j_1}.(\dotsm (b_{j_m}.v))) $$

(Roughly speaking: The basis elements of $\mathfrak{h}$ just do what they have always done, and the “new” basis elements are just put in front. Just as expected from a formal and general construction.)

Example: Let $\mathfrak{g} = \mathfrak{gl}_2(\mathbb{C})$ and $\mathfrak{h} = \mathfrak{sl}_2(\mathbb{C})$, and let $(e,h,f)$ be the basis of $\mathfrak{sl}_2(\mathbb{C})$ as above. We can extend this basis to a basis $(b,e,h,f)$ of $\mathfrak{gl}_2(\mathbb{C})$ by choosing $$ b = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}. $$ Notice that this basis of $\mathfrak{gl}_2(\mathbb{C})$ is already totally ordered by the way they are ordered in the tupel $(e,h,f,b)$. A basis of $\mathcal{U}(\mathfrak{gl}_n(\mathbb{C}))$ is now given by the monomials $b^k e^l h^m f^n$ with $k,l,m,n \in \mathbb{N}$, and a basis of $\mathcal{U}(\mathfrak{sl}_2(\mathbb{C}))$ is given by the monomials $e^l h^m f^n$ with $l,m,n \in \mathbb{N}$.

Consider the natural representation $V = \mathbb{C}^2$ of $\mathfrak{sl}_2(\mathbb{C})$, i.e. $x.v = x \cdot v$ for all $x \in \mathfrak{sl}_2(\mathbb{C})$ and $v \in \mathbb{C}^2$. Then the induced $\mathcal{U}(\mathfrak{gl}_2(\mathbb{C}))$-module structure on $\mathrm{Ind}_\mathfrak{h}^\mathfrak{g}(V)$ is given by $$ (b^k e^l h^m f^n) . (1 \otimes v) = b^k \otimes (e^l \cdot h^m \cdot f^n \cdot v) \quad \text{for all $v \in \mathbb{C}^2$}. $$

Notice that we we do not get the natural representation of $\mathfrak{gl}_2(\mathbb{C})$, as $b$ does not act by the identity, but rather via $b.(1 \otimes v) = b \otimes v$ for all $v \in \mathbb{C}^2$.

We are now nearly finished: We have constructed a $\mathcal{U}(\mathfrak{g})$-module structure on $\mathrm{Ind}_\mathfrak{h}^\mathfrak{g}(V)$, which by restriction of the multiplication to $\mathfrak{g} \subseteq \mathcal{U}(\mathfrak{g})$ now correspond to a representation of $\mathfrak{g}$. This representation is given by $$ x . ((x_1 \dotsm x_n) \otimes v) = (x \cdot x_1 \dotsm x_n) \otimes v. \quad \text{for all $x, x_1, \dotsc, x_n \in \mathfrak{g}$, $v \in V$}, $$ a special case of the previous formulae.

I hope this helps.

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  • $\begingroup$ This is incredibly helpful, thank you very much. $\endgroup$ – ChickenSocks Jan 3 '16 at 7:49
  • $\begingroup$ One quick question though: in your statement of the Poincaré-Birkhoff-Witt theorem the basis of U($\mathfrak{g})$ should be given by $x_{i_1}^{\epsilon_1}\cdots x_{i_n}^{\epsilon_n}$ for $\epsilon_i>0$, right? Not just $x_{i_1}\cdots x_{i_n}$? $\endgroup$ – ChickenSocks Jan 6 '16 at 1:38
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    $\begingroup$ Oh, you wrote that the indices $i_j$ could be equal, so I guess it amounts to the same thing. Nevermind! $\endgroup$ – ChickenSocks Jan 6 '16 at 1:44
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    $\begingroup$ I added it as a remark; it’s not like I wasn’t confused about this on some occasions. $\endgroup$ – Jendrik Stelzner Jan 6 '16 at 18:13

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