Reductio ad Abstractus

April 29, 2013

Cohomology of Lie Algebras (Part 3)

So far we’ve defined the cohomology groups $H^n(\mathfrak{g},A)$ for a $\mathfrak{g}$ -module $A$ , and we have computed $H^0 = A^{\mathfrak{g}}$ and $H^1 = \mathrm{Der}(\mathfrak{g},A)/\mathrm{Ider}(\mathfrak{g},A)$ . Since the rest of the theory quickly becomes very dense with computations and technical details, we’ll only sketch a brief overview of what’s going on.

Recall that to compute cohomology of a $\mathfrak{g}$ -module, we need to find a $\mathfrak{g}$ -projective resolution of the base field $K$ ; but so far we don’t have a standard procedure for constructing such a resolution. The idea is to take the vector space $V$ underlying $\mathfrak{g}$ , and to consider its nth exterior power $E_n V$ . For convenience, let us write $\langle x_1,\ldots,x_n \rangle$ for the product $x_1 \wedge \ldots \wedge x_n$ . Denote by $C_n$ ( $n = 0,1,\ldots$ ) the $\mathfrak{g}$ -module $U \mathfrak{g} \otimes_K E_n V$ . It may be shown that differentials $d_n: C_n \rightarrow C_{n-1}$ may be defined such that

$\cdots \rightarrow C_n \xrightarrow{d_n} C_{n-1} \rightarrow \cdots \rightarrow C_1 \xrightarrow{d_1} C_0 \xrightarrow{\epsilon} K$

is a $\mathfrak{g}$ -projective resolution of $K$ . Notice that $C_0 = U \mathfrak{g}$ and $\epsilon:C_0 \rightarrow K$ is just the augmentation. Since $E_n V$ is $K$ -free, it is clear that each $C_n$ is $\mathfrak{g}$ -free. Explicitly, the maps are given by

$\displaystyle \begin{aligned} d_n \langle x_1, \ldots, x_n \rangle = \sum_{i=1}^n (-1)^{i+1} x_i \otimes \langle x_1, \ldots, \hat{x_i}, \ldots, x_n \rangle \\ + \sum_{1 \leq i < j \leq n} (-1)^{i+j} \langle [x_i,x_j], x_1, \ldots, \hat{x_i}, \ldots, \hat{x_j}, \ldots, x_n \rangle \end{aligned}$

Notice the following important corollary for finite-dimensional Lie algebras:

Corollary. Let $\mathfrak{g}$ be a Lie algebra of dimension $n$ over $K$ . Then for any $\mathfrak{g}$ -module $A$ , $H^k(\mathfrak{g},A) = 0$ for $k \geq n+1$ .

Indeed, for $k \geq n+1$ we have $E_k V = 0$ . $\square$

From here on out, we shall only be concerned with finite-dimensional Lie algebras $\mathfrak{g}$ over a field $K$ of characteristic $0$ . Also, $A$ will denote a finite-dimensional $\mathfrak{g}$ -module.

Now, to any $\mathfrak{g}$ -module $A$ we can define an associated bilinear form by the following procedure: given the structure map $\rho:\mathfrak{g} \rightarrow L(\mathrm{End}_K A)$ , define $\beta:\mathfrak{g} \times \mathfrak{g} \rightarrow K$ by

$\beta(x,y) = \mathrm{Tr}(\rho(x) \circ \rho(y)), \quad x,y \in \mathfrak{g}$

Trivially, this form is bilinear and symmetric. In the special case where $A = \mathfrak{g}$ is regarded as a $\mathfrak{g}$ -module, the associated form is called the Killing form of $\mathfrak{g}$ . Thus the Killing form is given by $\beta(x,y) = \mathrm{Tr}(\mathrm{ad}(x) \circ \mathrm{ad}(y))$ .

We say that a Lie algebra $\mathfrak{g}$ is semi-simple if its only abelian ideal is $\{0\}$ .

Our goal is to prove the following theorem, due to Weyl: every finite-dimensional module $A$ over a semi-simple Lie algebra $\mathfrak{g}$ is a direct sum of simple $\mathfrak{g}$ -modules. We will need a key theorem from the theory of semi-simple Lie algebras, which is closely related to Cartan’s criterion for solvability of Lie algebras.

Theorem. Let $\mathfrak{g}$ be semi-simple (over a field of characteristic $0$ ) and let $A$ be a $\mathfrak{g}$ -module. If the structure map $\rho$ is injective, then the associated bilinear map $\beta$ corresponding to $A$ is non-degenerate. $\square$

Corollary. The Killing form of a semi-simple Lie algebra $\mathfrak{g}$ is non-degenerate.

Indeed, the structure map $\mathrm{ad}:\mathfrak{g} \rightarrow L(\mathrm{End}_K \mathfrak{g})$ has the center of $\mathfrak{g}$ as kernel; since the center is an abelian ideal, it is trivial. Hence $\mathrm{ad}$ is injective. $\square$

Corollary. Let $\mathfrak{a}$ be an ideal in the semi-simple algebra $\mathfrak{g}$ . Then there exists an ideal $\mathfrak{b}$ such that $\mathfrak{g} = \mathfrak{a} \oplus \mathfrak{b}$ , as Lie algebras.

Define $\mathfrak{b}$ to be the orthogonal complement of $\mathfrak{a}$ with respect to the Killing form $\beta$ . As is well-known, since we are working with finite-dimensional objects over a field, the non-degenerescence over $\beta$ implies that $\mathfrak{g}$ is the orthogonal direct sum of $\mathfrak{a}$ and $\mathfrak{b}$ . To see that $\mathfrak{b}$ is an ideal, let $x \in \mathfrak{g}, b \in \mathfrak{b}, a \in \mathfrak{a}$ . We have $\beta(a,[x,b]) = \beta([a,x],b) = \beta(a',b) = 0$ , where $[a,x] = a' \in \mathfrak{a}$ . Hence $b \in \mathfrak{b} \implies [x,b] \in \mathfrak{b}$ , and $\mathfrak{b}$ is an ideal. $\square$

Corollary. If $\mathfrak{g}$ is semi-simple, then every ideal $\mathfrak{a}$ in $\mathfrak{g}$ is semi-simple also.

By the corollary above, we have $\mathfrak{g} = \mathfrak{a} \oplus \mathfrak{b}$ . This implies that any ideal $\mathfrak{a'}$ in $\mathfrak{a}$ is also an ideal in $\mathfrak{g}$ . In particular, if $\mathfrak{a'}$ is abelian, then $\mathfrak{a'} = {0}$ . $\square$

Proposition. Let $A$ be a simple module over the semi-simple Lie algebra $\mathfrak{g}$ with non-trivial $\mathfrak{g}$ -action. Then $H^q(\mathfrak{g},A) = 0$ for all $q$ .

The idea is to consider the complement $\mathfrak{h}$ of the kernel $\mathfrak{h'}$ of the structure map $\rho:\mathfrak{g} \rightarrow A$ ; we have that $\mathfrak{h}$ is non-zero because $A$ is non-trivial. Since $\mathfrak{h}$ is semi-simple by the corollary above, and since $\rho$ restricted to $\mathfrak{h}$ is injective, the associated bilinear form $\beta$ is non-degenerate.

Pick bases $(e_i)$ and $(e_i ')$ of $\mathfrak{h}$ such that $\beta(e_i, e_j ') = \delta_{ij}$ . Take the standard resolution of $K$ described above. Construct $\mathfrak{g}$ -module homomorphisms $t_{C_n}:C_n \rightarrow C_n$ via $t_{C_n}(x) = \sum_i e_i \cdot (e_i ' \cdot x)$ . It can be shown that the chain map $\tau$ defined by the maps $t_{C_n}$ is homotopic to the zero map. However, by definition, we must have $\phi \circ t_A = t_{C_q} \circ \phi$ , i.e. $t_A^*(\phi) = (t_{C_q})_*(\phi)$ , for any $\mathfrak{g}$ -module homomorphism $\phi:C_q \rightarrow A$ . Hence, the map $H^q(\mathfrak{g},A) \rightarrow H^q(\mathfrak{g},A)$ induced by $t_A$ is the same as the map induced by $\tau$ , which is the zero map.

Now, since $A$ is simple, $t_A$ is either the zero map or an automorphism. But it cannot be the zero map, since the trace of $t_A$ equals $\sum_i \beta(e_i,e_i ') \neq 0$ by construction (recall that we are in characteristic $0$ ). Thus, $H^q(\mathfrak{g},A) = 0$ for all $q$ . $\square$

We need a final intermediate result, known as the first Whitehead lemma.

Proposition. Let $\mathfrak{g}$ be semi-simple. Then $H^1(\mathfrak{g},A) = 0$ .

Suppose instead that there exists a $\mathfrak{g}$ -module $A$ such that $H^1(\mathfrak{g},A) \neq 0$ . Then, there is such a $\mathfrak{g}$ -module $A$ with minimal $K$ -dimension. If $A$ is not simple, then there is a proper submodule $0 \neq A' \subset A$ . Consider the short exact sequence $0 \rightarrow A' \rightarrow A \rightarrow A/A' \rightarrow 0$ and the associated long exact cohomology sequence

$\cdots \rightarrow H^1(\mathfrak{g},A') \rightarrow H^1(\mathfrak{g},A) \rightarrow H^1(\mathfrak{g},A/A') \rightarrow \cdots$

Since the $K$ -dimensions of the the submodule and quotient module of $A$ are strictly smaller than that of $A$ , we must have $H^1(\mathfrak{g},A') = H^1(\mathfrak{g},A/A') = 0$ by minimality of $A$ . Hence $H^1(\mathfrak{g},A) = 0$ , a contradiction. Thus $A$ has to be simple. But then, by the proposition above, $\mathfrak{g}$ must act trivially on $A$ . Then by the last proposition in part 2, we have $H^1(\mathfrak{g},A) = \mathrm{Hom}_K(\mathfrak{g}_{ab},A)$ . Now consider

$[\mathfrak{g},\mathfrak{g}] \hookrightarrow \mathfrak{g} \twoheadrightarrow \mathfrak{g}_{ab}$

By the corollary above, the ideal $[\mathfrak{g},\mathfrak{g}]$ has a complement in $\mathfrak{g}$ , which must be isomorphic to $\mathfrak{g}_{ab}$ for dimensionality reasons. In particular, it must be abelian, hence $\mathfrak{g}_{ab} = 0$ ( $\mathfrak{g}$ is semi-simple). Thus $H^1(\mathfrak{g},A) \cong \mathrm{Hom}_K(\mathfrak{g}_{ab},A) = 0$ , which contradicts our initial assumptions. $\square$

We are finally read to prove Weyl’s decomposition theorem of a module over a semi-simple Lie algebra:

Theorem. (Weyl) Every (finite-dimensional) module $A$ over a semi-simple Lie algebra $\mathfrak{g}$ is a direct sum of simple $\mathfrak{g}$ -modules.

Using induction on the $K$ -dimension of A, it suffices to show that every non-trivial submodule $0 \neq A' \subset A$ is a direct summand in $A$ . Consider the short exact sequence $0 \rightarrow A' \rightarrow A \rightarrow A'' \rightarrow 0$ and the induced sequence

$0 \rightarrow \mathrm{Hom}_K(A'',A') \rightarrow \mathrm{Hom}_K(A,A') \rightarrow \mathrm{Hom}_K(A',A') \rightarrow 0$

which is exact since we are dealing with vector spaces. Each of those vector spaces are finite-dimensional and can be given a $\mathfrak{g}$ -module structure by the following procedure. If $B,C$ are $\mathfrak{g}$ -modules, then put $(x \cdot f)(b) = x \cdot f(b) - f(x \cdot b)$ . With this understanding, the sequence above becomes a short exact sequence of $\mathfrak{g}$ -modules. Note that the invariant elements in $\mathrm{Hom}_K(B,C)$ are precisely the $\mathfrak{g}$ -module homomorphisms from $B$ to $C$ . Now consider the long exact sequence arising from the above sequence:

$\begin{aligned} 0 & \rightarrow H^0(\mathfrak{g},\mathrm{Hom}_K(A'',A')) \rightarrow H^0(\mathfrak{g},\mathrm{Hom}_K(A,A')) \\ & \rightarrow H^0(\mathfrak{g},\mathrm{Hom}_K(A',A')) \rightarrow H^1(\mathfrak{g},\mathrm{Hom}_K(A'',A')) \rightarrow \cdots \end{aligned}$

By the proposition above, the last term vanishes. Passing to the interpretation of $H^0$ as the group of invariant elements, we obtain an epimorphism

$\mathrm{Hom}_{\mathfrak{g}}(A,A') \twoheadrightarrow \mathrm{Hom}_{\mathfrak{g}}(A',A')$

It follows that there is a $\mathfrak{g}$ -module homomorphism $A \rightarrow A'$ inducing the identity in $A'$ ; hence our short exact sequence $0 \rightarrow A' \rightarrow A \rightarrow A'' \rightarrow 0$ splits. $\square$

April 25, 2013

Cohomology of Lie Algebras (Part 2)

Recall from the last post that if we are given a Lie algebra $\mathfrak{g}$ , we may define the notion of a $\mathfrak{g}$ -module, which, by the adjunction $\eta:U \dashv L$ , is essentially the same thing as a module over the universal enveloping algebra $U \mathfrak{g}$ of $\mathfrak{g}$ . With this in mind, we shall write $\mathrm{Hom}_{\mathfrak{g}}$ , $\mathrm{Ext}_{\mathfrak{g}}^n$ , etc., in place of $\mathrm{Hom}_{U \mathfrak{g}}$ , $\mathrm{Ext}_{U \mathfrak{g}}^n$ , etc.

Given a Lie algebra $\mathfrak{g}$ over $K$ and a $\mathfrak{g}$ -module $A$ , we define the nth cohomology group of $\mathfrak{g}$ with coefficients in $A$ by

$H^n(\mathfrak{g},A) = \mathrm{Ext}_{\mathfrak{g}}^n(K,A), \quad n=0,1,\ldots$

where $K$ is to be regarded as a trivial $\mathfrak{g}$ -module. Thus, to compute the cohomology groups, one should first take a $\mathfrak{g}$ -projective resolution of the base field $K$ , apply the contravariant functor $\mathrm{Hom}_{\mathfrak{g}}(-,A)$ to the deleted resolution, and take cohomology.

Let us first compute $H^0(\mathfrak{g},A)$ , for any $\mathfrak{g}$ -module $A$ . Since $\mathrm{Hom}(-,A)$ is a left-exact functor, we have $\mathrm{Ext}_{\mathfrak{g}}^0 \cong \mathrm{Hom}_{\mathfrak{g}}(K,A)$ . But by $K$ -linearity, any $\phi \in \mathrm{Hom}_{\mathfrak{g}}(K,A)$ is entirely determined by its action on $1 \in K$ . Also, since $\phi$ is a $\mathfrak{g}$ -module homomorphism, if $\phi(1) = a$ we must have

$x \cdot a = x \cdot \phi(1) = \phi(x \cdot 1) = 0$

for all $x \in \mathfrak{g}$ . Thus, we conclude that

$H^0(\mathfrak{g},A) = \{a \in A \mid x \cdot a = 0, \quad \forall x \in \mathfrak{g}\}$

We call this the subspace of invariants of $A$ and denote it by $A^{\mathfrak{g}}$ .

In order to exhibit the nature of $H^1(\mathfrak{g},A)$ , we introduce the notion of Lie algebra derivations. We say that a $K$ -linear map $d:\mathfrak{g} \rightarrow A$ is a derivation if

$d([x,y]) = x \cdot d(y) - y \cdot d(x), \quad x,y \in \mathfrak{g}$

We can endow the set $\mathrm{Der}(\mathfrak{g},A)$ of all derivations $d:\mathfrak{g} \rightarrow A$ with a $K$ -vector space structure in the natural way. Denote by $\mathrm{Ider}(\mathfrak{g},A)$ the subspace of all inner derivations, that is, derivations of the form $d_a(x) = x \cdot a$ . Also recall that $I \mathfrak{g}$ is the augmentation ideal of $U \mathfrak{g}$ , and corresponds to the “non-constant” elements of $U \mathfrak{g}$ , $U \mathfrak{g} \cong K \oplus I \mathfrak{g}$ .

Theorem. The functor $\mathrm{Der}(\mathfrak{g},-)$ is represented by the $\mathfrak{g}$ -module $I \mathfrak{g}$ . In other words, for any $\mathfrak{g}$ -module $A$ there is a natural isomorphism between the $K$ -vector spaces $\mathrm{Der}(\mathfrak{g},A)$ and $\mathrm{Hom}_{\mathfrak{g}}(I \mathfrak{g},A)$ .

The proof is straightforward. Given a derivation $d$ , define $f'_d:T \mathfrak{g} \rightarrow A$ by sending $K$ into zero and setting

$f'_d(x_1 \otimes \cdots \otimes x_n) = x_1 \cdot (x_2 \cdot \, \cdots \, \cdot ( x_{n-1} \cdot d x_n) \ldots )$

Then since $d$ is a derivation and $A$ is a $\mathfrak{g}$ -module, $f'_d$ vanishes on all elements of the form

$t_1 \otimes (x \otimes y - y \otimes x - [x,y]) \otimes t_2$

where $t_1,t_2 \in T \mathfrak{g}, x,y \in \mathfrak{g}$ . Thus $f'_d$ induces a $\mathfrak{g}$ -module homomorphism $f_d:I \mathfrak{g} \rightarrow A$ , as required. Conversely, if $f:I \mathfrak{g} \rightarrow A$ is given, extend $f$ to $U \mathfrak{g}$ by setting $f(K) = 0$ , and put $d_f = f \circ i:\mathfrak{g} \rightarrow A$ . Then it’s easy to see that $d_f$ is a derivation, $f_{d_f} = f$ , $d_{f_d} = d$ , and that the map $f \mapsto d_f$ is $K$ -linear. $\square$

Now, taking the obvious $\mathfrak{g}$ -free presentation of $K$

$I \mathfrak{g} \hookrightarrow U \mathfrak{g} \twoheadrightarrow K$

we obtain, given a $\mathfrak{g}$ -module $A$ ,

$H^1 (\mathfrak{g},A) = \mathrm{coker}(\mathrm{Hom}_{\mathfrak{g}}(U \mathfrak{g},A) \rightarrow \mathrm{Hom}_{\mathfrak{g}}(I \mathfrak{g},A))$

Thus, the first cohomology group is isomorphic to the space of derivations from $\mathfrak{g}$ into $A$ , modulo those that arise from $\mathfrak{g}$ -module homomorphisms $f:U \mathfrak{g} \rightarrow A$ . But if $f(1_{U \mathfrak{g}}) = a$ , then $d_f(x) = x \cdot d_f(1_{U \mathfrak{g}}) = x \cdot a$ , so that these are precisely the inner derivations.

Proposition. $H^1(\mathfrak{g},A) \cong \mathrm{Der}(\mathfrak{g},A)/\mathrm{Ider}(\mathfrak{g},A)$ . Moreover, if $A$ is a trivial $\mathfrak{g}$ -module, then $H^1(\mathfrak{g},A) \cong \mathrm{Hom}_K(\mathfrak{g}_{ab},A)$ .

Only the second assertion remains to be proved. By $\mathfrak{g}_{ab}$ ,we mean the abelianization of $\mathfrak{g}$ , i.e. the quotient of $\mathfrak{g}$ by the ideal generated by the commutators. Since $A$ is trivial, there are no non-trivial inner derivations, and a derivation $d:\mathfrak{g} \rightarrow A$ is simply a Lie algebra homomorphism, where $A$ is regarded as an abelian Lie algebra:

$d([x,y]) = x \cdot dy - y \cdot dx = 0 = [dx,dy]$

In other words, a derivation is the same thing as a $K$ -linear map $\mathfrak{g}_{ab} \rightarrow A$ . $\square$

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April 13, 2013

Cohomology of Lie Algebras (Part 1)

In this post and subsequent ones, we shall use the theory of derived functors to deduce important facts about Lie algebras. Let $K$ be a field. Recall that a Lie Algebra $\mathfrak{g}$ over $K$ is a $K$ -vector field endowed with a special bilinear product $[-,-]:\mathfrak{g} \times \mathfrak{g} \rightarrow \mathfrak{g}$ . This Lie bracket is assumed to satisfy the following identities:

$[x,x]=0, \quad x \in \mathfrak{g}$
$[[x,y],z] + [[y,z],x] + [[z,x],y] = 0, \quad x,y,z \in \mathfrak{g}$

Lie algebras arise as algebraic objects encoding the local structure of Lie groups (which are topological groups endowed with a smooth structure). More precisely, the Lie algebra of a Lie group is its tangent space at the identity. We shall not venture deeper into this territory; instead we turn our focus to the algebraic aspects of the theory.

A Lie algebra homomorphism is a linear map that preserves the Lie bracket. A Lie subalgebra $\mathfrak{h} \subset \mathfrak{g}$ is a subspace closed under the Lie bracket. We say that the subalgebra $\mathfrak{h}$ is a Lie ideal of $\mathfrak{g}$ if $[x,y] \in \mathfrak{h}$ for all $x \in \mathfrak{g}, y \in \mathfrak{h}$ . The Lie algebra $\mathfrak{g}$ is called abelian if its Lie bracket vanishes everywhere. Thus any vector space may be trivially regarded as an abelian Lie algebra.

Now, to any (associative, unital) $K$ -algebra $\Lambda$ we can associate (functorially) a Lie algebra $L\Lambda$ on the same underlying vector space as $\Lambda$ , by defining $[x,y] = xy - yx$ . It is easy to see that the Lie algebra axioms for $L\Lambda$ are satisfied. Note, however, that the Lie bracket is not associative in general. Thus, we get a functor $L$ from the category $Alg$ of $K$ -algebras to the category $Lie$ of Lie algebras. Furthermore, this functor possesses a left adjoint $U$ , which we proceed to describe.

Given a Lie algebra $\mathfrak{g}$ , we define the universal enveloping algebra $U \mathfrak{g}$ of $\mathfrak{g}$ to be the quotient of the tensor algebra $T \mathfrak{g}$ by the ideal $I$ generated by the elements of the form $x \otimes y - y \otimes x - [x,y]$ , for all $x,y \in \mathfrak{g}$ . Thus

$U \mathfrak{g} = T \mathfrak{g} / (x \otimes y - y \otimes x - [x,y])$

Theorem. The functors $U$ and $L$ form an adjoint pair ( $U \dashv L$ for short).

To show that $U$ is left adjoint to $L$ , recall that we need to find a natural equivalence $\eta:Alg(U-,-) \tilde{\rightarrow} Lie(-,L-)$ of bifunctors $Lie^{op} \times Alg \rightarrow \mathfrak{S}$ . More precisely, given a Lie algebra homomorphism $f:\mathfrak{g} \rightarrow L \Lambda$ , we need to find a unique $K$ -algebra homomorphism $f_1:U \mathfrak{g} \rightarrow \Lambda$ to which it naturally corresponds. However, viewing $f$ as a linear mapping on the underlying $K$ -vector spaces, the universal property of the tensor algebra (as the free object over its underlying vector space) gives us a unique $K$ -algebra homomorphism $f_0:T \mathfrak{g} \rightarrow \Lambda$ extending $f$ ; it is defined by the rule

$f_0(x_1 \otimes \cdots \otimes x_n) = f(x_1) \cdot \, \cdots \, \cdot f(x_n)$

Moreover, this map vanishes on our ideal $I$ : on any generator we have

$\begin{aligned}f_0(x \otimes y - y \otimes x - [x,y]) &= f(x)f(y) - f(y)f(x) - f([x,y])\\ &= f(x)f(y) - f(y)f(x) - [f(x),f(y)]\\ &= 0 \end{aligned}$

Thus $f_0$ descends to a unique $K$ -algebra homomorphism $f_1:U \mathfrak{g} \rightarrow \Lambda$ , as required. Naturality follows easily from the explicit definitions of the maps involved. $\square$

Note that we have a canonical $K$ -vector space map $i_{\mathfrak{g}}$ given by the composition $\mathfrak{g} \hookrightarrow T \mathfrak{g} \twoheadrightarrow U \mathfrak{g}$ . This map gives rise to a Lie algebra homomorphism $i_{\mathfrak{g}}:\mathfrak{g} \rightarrow LU \mathfrak{g}$ ; in fact, $i$ is exactly the unit of the adjunction between $U$ and $L$ , since it corresponds via $\eta$ to the identity homomorphism $1_{U \mathfrak{g}}$ . In terms of this map, the equivalence above is explicitly given by the formula $\eta(\phi) = L\phi \circ i_{\mathfrak{g}}$ , where $\phi$ is some algebra homomorphism $U \mathfrak{g} \rightarrow \Lambda$ .

We shall accept without (tedious) proof the following very important

Theorem. (Birkhoff-Witt) Let $\{e_i\}_{i \in J}$ be a $K$ -basis of $\mathfrak{g}$ indexed by a totally ordered set $J$ . Let $I = (i_1,i_2,\ldots,i_k)$ denote an increasing sequence of elements in $J$ , and define $e_I = e_{i_1} e_{i_2} \cdots e_{i_k} \in U \mathfrak{g}$ to be the projection of $e_{i_1} \otimes e_i{_2} \otimes \cdots \otimes e_{i_k} \in T \mathfrak{g}$ . Then the elements $e_I$ corresponding to all finite increasing sequences $I$ (including the empty one) form a $K$ -basis for $U \mathfrak{g}$ . $\square$

Corollary. The unit $i_{\mathfrak{g}}:\mathfrak{g} \rightarrow LU \mathfrak{g}$ is an embedding. Thus, every Lie algebra $\mathfrak{g}$ over $K$ is isomorphic to a Lie subalgebra of a Lie algebra of the form $L \Lambda$ for some $K$ -algebra $\Lambda$ . $\square$

Now, in a manner similar to how we define modules over rings, let us say that a $K$ -vector space $A$ is a (left) $\mathfrak{g}$ -module if we endow it with a Lie algebra homomorphism $\rho:\mathfrak{g} \rightarrow L(End_K A)$ . We may therefore think of $\mathfrak{g}$ as acting on $A$ on the left and write $x \cdot a$ for $\rho(x)(a)$ . Then $A$ is a $\mathfrak{g}$ -module if $x \cdot a$ is $K$ -linear in $x$ and $a$ and

$[x,y] \cdot a = x \cdot (y \cdot a) - y \cdot (x \cdot a), \quad x,y \in \mathfrak{g}, a \in A$

By the universal property of $U \mathfrak{g}$ (as stated above in terms of an adjunction), a structure map $\rho:\mathfrak{g} \rightarrow L(End_K A)$ gives rise to a unique algebra homomorphism $\rho_1 = \eta^{-1}(\rho):U \mathfrak{g} \rightarrow End_K A$ . Thus, we can view $A$ as a left $U \mathfrak{g}$ -module. Conversely, any left $U \mathfrak{g}$ -module $A$ with structure map $\sigma:U \mathfrak{g} \rightarrow End_K A$ may be viewed as a left $\mathfrak{g}$ -module by setting $\rho = L\sigma \circ i_{\mathfrak{g}} = \eta(\sigma)$ . Thus, the notions of $\mathfrak{g}$ -module and $U \mathfrak{g}$ module effectively coincide, by means of the adjunction $\eta:U \dashv L$ .

It is important to note that a Lie algebra $\mathfrak{g}$ itself may be regarded as a $\mathfrak{g}$ -module. The structure map $\mathrm{ad}:\mathfrak{g} \rightarrow L(End_K \mathfrak{g})$ is called the adjoint endomorphism and is defined by

$(\mathrm{ad} x)(z) = [x,z], \quad x,z \in \mathfrak{g}$

A $\mathfrak{g}$ -module homomorphism $\phi:A \rightarrow B$ between $\mathfrak{g}$ -modules $A,B$ is simply a $K$ -linear map satisfying $\phi(x \cdot a) = x \cdot \phi(a)$ , naturally enough. By the correspondence above, a $\mathfrak{g}$ -module homomorphism is essentially the same thing as a $U \mathfrak{g}$ -module homomorphism.

Finally, let us say that a $\mathfrak{g}$ -module $A$ is trivial if its structure map is trivial, i.e. if $x \cdot a = 0$ for all $x \in \mathfrak{g}$ . Clearly, we can identify a $K$ -vector space with a trivial $\mathfrak{g}$ -module for any Lie algebra $\mathfrak{g}$ . The algebra homomorphism $\epsilon:U \mathfrak{g} \rightarrow K$ associated with the trivial structure map of $K$ is called the augmentation of $U \mathfrak{g}$ . It only leaves the copy of $K$ sitting inside $U \mathfrak{g}$ intact. The kernel $I \mathfrak{g}$ of $\epsilon$ is called the augmentation ideal of $\mathfrak{g}$ . It is simply the ideal of $U \mathfrak{g}$ generated by $i_{\mathfrak{g}}(\mathfrak{g})$ .

Now, with all these definitions out the way, let us see if we can finally obtain some tangible facts on Lie algebras.

March 15, 2013

Of Sets, Bijections and Cardinalities

Today we are staying inside the category of sets! In mathematics, one of the most elementary and important ideas is that of a bijection. Let’s say we have two sets $A,B$ of objects. We would like to understand how to compare the size of these sets, even if they are very large (or even infinite!). Well, let’s take a mapping $f:A \rightarrow B$ , that this, a process that takes any element $a \in A$ and sends it to an element $f(a) \in B$ .

We say that $f$ is injective if it maps distinct objects to distinct objects. Thus, intuitively, an injective mapping cannot map a set to a smaller set:

Similarly, we say that $f$ is surjective if every object $b$ in the arrival set $B$ has a preimage, that is, an $a \in A$ such that $f(a) = b$ . Intuitively, a surjective mapping cannot map a set to a larger set:

Combining these two notions, we arrive at the definition of a bijective mapping: we say that $f:A \rightarrow B$ is bijective if it is both injective and surjective. By virtue of our intuition above, a bijection should precisely capture the notion of “equality of size”:

This really corresponds to the way we should think about two sets being “essentially the same” (or, to be precise, isomorphic): even if we don’t explicitly know what their elements are, if we can match their elements by finding a one-to-one correspondence between them, then clearly the two sets should be considered equivalent. We can then speak of equivalence classes (or isomorphism classes) of sets: in such a class we regroup all the sets that are in bijective correspondence. In the picture above, $A$ and $B$ are in the same equivalence class.

We associate to every equivalence class of sets (or any set $A$ contained in such an equivalence class) a “number” $|A|$ , which is more formally referred to as a cardinal number. This number is precisely the measure of the size of the sets contained in our class. If a set is finite and possesses $n$ elements, we say that its cardinality is $n$ . We can also associate such numbers to classes of infinite sets; we call them transfinite cardinals. So we have a list

$0,1,2,3,\ldots,n,\ldots,\aleph_0,\aleph_1,\aleph_2,\ldots,\aleph_\alpha,\ldots$

of cardinals. The first transfinite cardinal is denoted by $\aleph_0$ (“aleph naught”) and corresponds to the size of the set of natural numbers $\mathbb{N} = \{0,1,2,3,\ldots\}$ ; it can be shown that it really is the smallest infinite cardinal. But what does it mean for a cardinal to be smaller than another?

We can order the cardinals as follows: let us write $|A| \leq |B|$ if and only if there exists an injective mapping $f:A \rightarrow B$ . This corresponds to our intuition above, and is also compatible with the standard ordering of the natural numbers:

A classic theorem of Cantor, Bernstein and Schroeder affirms that if $|A| \leq |B|$ and $|B| \leq |A|$ , then $|A| = |B|$ (in that case $A$ and $B$ are in bijective correspondence). Also, the axiom of choice assures us that we always have $|A| \leq |B|$ or $|B| \leq |A|$ ; thus, this ordering is very natural.

However, how do we know that the list of transfinite cardinals never stops? In other words, is it possible to explicitly construct infinite sets that grow larger and larger? The answer is yes, and easily! Start with a set $A_0$ of cardinality $\aleph_0,$ for instance, the natural numbers $A_0 = \mathbb{N} = \{0,1,2,3,\ldots\}$ . Now, let $A_1 = P(A_0)$ be the power set of $A_0$ , that is, the set of all subsets of $A_0$ . Clearly $|A_0| \leq |A_1|$ , since we can send any element $x \in A_0$ to the subset $\{x\} \in A_1$ , and that mapping is easily seen to be injective. However, I affirm that there exists no surjective mapping from $A_0$ to $A_1$ . We prove this by contradiction: let $f:A_0 \to A_1$ be the putative mapping. Now define $B = \{ x \in A_0 \mid x \not\in f(x) \}$ . I affirm that $B$ is not in the image of $f$ , contradicting the surjectivity of $f$ . Indeed, if it were, then there would exist an element $x \in A_0$ such that $f(x) = B$ . But this is impossible by definition of $B$ ! Hence, iterating this process, we arrive at an infinite sequence $A_0, A_1, A_2, \ldots$ of sets of strictly growing cardinality.

What this whole story tells is that there are many (in fact, infinitely many) “sizes” of infinity (!), and that it is possible to compare them. That, in itself, is quite a remarkable fact.

This story also tells us that to study mathematical objects, it is sometimes (actually, most of the time, as it turns out) beneficial to study the links between them and the properties of these links. In our simple scenario, a mapping $f:A \rightarrow B$ gives us a way to compare $A$ and $B$ . In a way, if we seek to understand some object $A$ , then the collection of all mappings from $A$ to some $B$ is, in itself, a very important object of study. The generalization of this notion leads to the concept of a category, which is a fundamental tool in the development and unification of modern mathematics.

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March 14, 2013

Additive Categories

In this post we investigate some interesting facts about additive categories. Recall that a zero object in some category $\mathfrak{C}$ is an element $Z$ that is both initial and terminal in $\mathfrak{C}$ . This means that for any object $A \in \mathfrak{C}$ , the hom-sets $\mathfrak{C}(A,Z)$ and $\mathfrak{C}(Z,A)$ possess exactly one element. In particular, given any $A,B \in \mathfrak{C}$ , there is a well-defined morphism $A \rightarrow Z \rightarrow B$ , called the zero morphism from $A$ to $B$ . Since zero objects are essentially unique (meaning that they are unique up to unique isomorphism), the zero morphism does not depend on the choice of the zero object $Z$ .

We seek to generalize the nice properties of $\mathfrak{Ab}$ (the category of abelian groups), or more generally $\mathfrak{M}_{\Lambda}$ (the category of $\Lambda$ -modules, for some unitary ring $\Lambda$ ). Note that we do have a zero object in $\mathfrak{M}_{\Lambda}$ , namely the zero module. This means that the zero morphism between any two modules is the usual zero module homomorphism. We can also adorn the hom-set $\mathfrak{M}_{\Lambda}(A,B)$ with an abelian group structure in the obvious way. Generalizing, we say that a category $\mathfrak{A}$ is preadditive if it has a zero object and if $\mathfrak{A}(A,B)$ has an abelian group structure, such that the composition is bilinear. This last condition simply means that $f(g+h) = fg + fh$ and $(f+g)h = fh + gh$ .

Here’s a nice way to think about this: for any category $\mathfrak{C}$ we always have a hom-set functor $\mathfrak{C}^{op} \times \mathfrak{C} \rightarrow \mathfrak{S}$ that maps $(A,B)$ to $\mathfrak{C}(A,B)$ . But if we restrict ourselves to preadditive categories $\mathfrak{A}$ , we get a functorial lifting

where $U$ is the forgetful functor.

Note that there is no ambiguity when we speak of the zero morphism in $\mathfrak{A}(A,B)$ : since $0 = 0(0+0) = 0 + 0$ , by cancelling zeroes we get that the categorical zero morphism is exactly the zero morphism in our abelian group.

Now, an additive category $\mathfrak{A}$ is a preadditive category in which any two objects have a product. Recall that a product of $A_i \in \mathfrak{A}$ is an object $\prod_i A_i$ , together with a family of morphisms $p_i:\prod_j A_j \rightarrow A_i$ (called the projection onto the i’th factor), which satisfy the following universal property (the illustration represents the case where we have two factors):

In other words, for every object $B$ and every family of morphisms $\phi_i:B \rightarrow A_i$ , there exists a unique morphism $\psi:B \rightarrow \prod_i A_i$ such that $\phi_i = p_i \psi$ .

This definition might lead one to think that coproducts might fail to exist in a general additive category. However, as we will now see, the axioms really are self-dual. The idea is to construct “canonical injections” by means of the zero and identity morphisms. In fact, the universal property of the product of $A_1,A_2 \in \mathfrak{A}$ guarantees us the existence of morphisms

$i_1 = \{1,0\}:A_1 \rightarrow A_1 \prod A_2$ and $i_2 = \{0,1\}:A_2 \rightarrow A_1 \prod A_2$

satisfying $1 = p_j i_j$ , $0 = p_j i_k \, (j \neq k)$ .

Lemma 1. We have $i_1 p_1 + i_2 p_2 = 1$ .

Indeed, we see that $p_1(i_1 p_1 + i_2 p_2) = p_1$ and $p_2(i_1 p_1 + i_2 p_2) = p_2$ , so that by the uniqueness property of the product, $i_1 p_1 + i_2 p_2 = 1$ . $\square$

Proposition 1. $\left( A_1 \prod A_2, (i_j) \right)$ is the coproduct of $A_1$ and $A_2$ in $\mathfrak{A}$ .

Given $\varphi_i:A_i \rightarrow B$ , define $\langle \varphi_1, \varphi_2 \rangle = \varphi_1 p_1 + \varphi_2 p_2:A_1 \prod A_2 \rightarrow B$ . We have $\langle \varphi_1, \varphi_2 \rangle i_1 = (\varphi_1 p_1 + \varphi_2 p_2)i_1 = \varphi_1$ and $\langle \varphi_1, \varphi_2 \rangle i_2 = (\varphi_1 p_1 + \varphi_2 p_2)i_2 = \varphi_2$ , as needed. For the uniqueness part, we invoke lemma 1: if $\theta:A_1 \prod A_2 \rightarrow B$ also satisfies $\varphi_j = \theta i_j$ , then $\theta = \theta 1 = \theta (i_1 p_1 + i_2 p_2) = \varphi_1 p_1 + \varphi_2 p_2 = \langle \varphi_1, \varphi_2 \rangle$ . $\square$

Thus, in an additive category, we have both finite products and coproducts, and they coincide. From now on, we shall write $A_1 \oplus A_2$ for the (co)product of $A_1$ and $A_2$ . An easy calculation shows that, given $A \xrightarrow{\{ \varphi, \psi \}} B \oplus C \xrightarrow{\langle \gamma,\delta \rangle} D$ , we have $\langle \gamma, \delta \rangle \{ \varphi, \psi \} = \gamma \varphi + \delta \psi$ . This has the following interesting corollary:

Corollary 1. The addition in the set $\mathfrak{A}(A,B)$ is determined by the additive category $\mathfrak{A}$ .

Indeed, if $\varphi_1, \varphi_2:A \rightarrow B$ , then $\varphi_1 + \varphi_2 = \langle \varphi_1, \varphi_2 \rangle \{ 1, 1 \}$ . $\square$

Thus, given the slightly stronger requirement that we work in an additive category, i.e. that our preadditive category possess finite products, not only can we functorially lift the hom-set functor to $\mathfrak{Ab}$ , but this lifting is unique!

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