### Sumários

##### The representations of sl(2,C) and sl(3,C)

3 Fevereiro 2022, 12:30 • John Huerta

**The representations of sl _{2}(**ℂ

**)**

Remember where we were: we fixed a nice basis of sl

_{2}(ℂ), consisting of matrices E, H and F, which satisfy:

[H, E] = 2E, [E,F] = H, [H, F] = -2F.

We decompose any irrep V into the eigenspaces of H:

V = ⊕

_{a}V

_{a}

_{}

In Lie theory, the eigenvalues occuring here have a special name: they are called the

**weights**of V. The eigenspaces V

_{a}are called

**weight spaces**. An eigenvector v in V

_{a}is called a

**weight vector**.

Last time we saw that if v is a weight vector in V

_{a}, then Ev is a weight vector in V

_{a+2 }, while Fv is a weight vector in V

_{a-2}. So we get this picture of the representation:

V = ... ⊕ V

_{a-2}⊕ V

_{a}⊕ V

_{a+2}⊕ ...

where H multiplies by the weight, E rasies the weight, and F lowers the weight. Because V is finite dimensional, this cannot go on forever, so there must be a largest weight in this sequence:

V = ... ⊕ V

_{a-2}⊕ V

_{a}⊕ V

_{a+2}⊕ ... ⊕ V

_{a_max}

The weight a_max is called the

**highest weight**, and a nonzero vector v in V

_{a_max}is called a

**highest weight vector**. A highest weight vector has the following property: if we try to raise the weight with E, we get 0: Ev = 0.

Similarly, there must be a lowest weight:

V = V

_{a_min}⊕ ... ⊕ V

_{a-2}⊕ V

_{a}⊕ V

_{a+2}⊕ ... ⊕ V

_{a_max}

and for any vector v in V

_{a_min}, we must have Fv = 0.

Now, if we start with a highest weight vector v in V

_{a_max}and keep lowering the weight with F, we will eventually get a vector in V

_{a_min}. Let us suppose this happens in m steps. Then m is a natural number such that:

F

^{m}v is nonzero, but F

^{m+1}v = 0.

**Proposition.**The vectors {v, Fv, F

^{2}v, ..., F

^{m}v} form a basis of V.

**Proof.**

These vectors are linearly independent because they are eigenvectors (weight vectors) with distinct eigenvalues (weights).

To show they span V, let W = span{v, Fv, F

^{2}v, ..., F

^{m}v}. Then W is a nozero subspace of V, and it is preserved by the action of E (which raises the weight), F (which lower the weight) and H (which multiplies by the weight). Thus, W is a nonzero subrepresentation of V. Because V is irreducible, we conclude W = V. QED.

In this basis, we know exactly how H acts:

H F

^{k}v = (a_max - 2k) v

and we know exactly how F acts

F F

^{k}v = F

^{k+1}v.

But it is less clear how E acts:

E F

^{k}v = ?

Let us derive a formula for the action E. First, we know Ev = 0. For EFv, we compute:

EFv = FEv + [E,F]v = 0 + Hv = a_max v.

And for EF

^{2}v, we compute:

EF

^{2}v = FEFv + [E,F]Fv = a_max Fv + HFv = (a_max + (a_max - 2)) Fv.

Continuing in this way, we can discover the pattern is:

EF

^{k+1}v = (a_max + (a_max - 2) + (a_max - 4) + ... + (a_max - 2k)) F

^{k}v.

This formula simplifies to:

EF

^{k+1}v = (k+1)(a_max - k) F

^{k}v.

We learn something magical when we apply this formula for k = m:

EF

^{m+1}v = 0 = (m+1)(a_max - m) F

^{m}v.

It vanishes because F

^{m}v is in the lowest weight space, so F

^{m+1}v = 0. But on the right hand side, the m+1 factor is nonzero, and the vector F

^{m}v is nonzero. So the only way for the right hand side to vanish is if:

a_max = m.

*Look at this.*The highest weight a_max is natural number! Specifically, it is exactly the number of steps we need to go from the highest weight vector v to the lowest.

**Theorem**. For each natural number m (including zero), there is a unique irreducible representation (V

^{(m)}, π

_{m}) of sl

_{2}(ℂ)

**with highest weight m.**

**Exercises.**You should prove that:

- (V
^{(0)}, π_{0}) is the trivial representation ℂ. - (V
^{(1)}, π_{1}) is the standard representation ℂ^{2}. - (V
^{(2)}, π_{2}) is the adjoint representation.

**The representations of sl**.

_{3}(ℂ)*NONE OF THE FOLLOWING WILL BE ON YOUR EXAM.*

We don't have time for much more. The theory for sl

_{3}(ℂ) starts the same way: choose a Cartan subalgebra h in sl

_{3}(ℂ). As with sl

_{3}(ℂ), we choose h to be the subalgebra of diagonal matrices. Now h is 2-dimensional:

h = span{ H

_{1}, H

_{2}}

where H

_{1}and H

_{2}are the matrices:

H

_{1}=

1 |
0 |
0 |

0 |
-1 |
0 |

0 |
0 | 0 |

and

H

_{2}=

0 |
0 | 0 |

0 |
1 | 0 |

0 |
0 | -1 |

As before, the Cartan subalgebra h is maximal abelian, and ad

_{H}is diagonalizable for all H: recall, from last lecture, that we have the formula ad

_{H}(E

_{ij}) = (a

_{i}- a

_{j}) E

_{ij}, where H is the diagonal matrix

a_{1} |
||

a_{2} |
||

a_{3} |

and E

_{ij}is the elementary matrix, with 1 in the ijth entry and 0 elsewhere.

The idea is again to apply

**Deep theorem.**Given any finite-dimensional complex representation (V, ∏) of sl

_{n}(ℂ), then ∏

_{H}is diagonalizable for all H in h.

We use this, analogously to sl

_{2}(ℂ), to write a representation V of sl

_{3}(ℂ) as the direct sum of weight spaces:

V = ⊕

_{a}V

_{a}

_{}

But now we meet a real difference. Because, there was only one H, and a weight was simply an eigenvalue of H. Now there are many Hs. What is a weight? What is a weight space?

**Definition.**A

**weight**a is a linear map a : h -> ℂ. A

**weight vector**v with weight a in a representation V is a vector v in V such that Hv = a(H) v, for all H in h. The weights with nonzero weight vectors in a representation V are called the

**weights of V**.

Weights are a generalization of eigenvalues, and weight vectors are a generalization of eigenvectors, that allow us to diagonalize all the H in h at once. And we really can diagonalize all the H in h simultaneously, because they commute!

To get a feel for the weights of a representation of sl

_{3}(ℂ), let us consider an example.

**Example.**The weights of the adjoint representation.

We already know the weight vectors - they're the elementary matrices E

_{ij}, at least for i not equal to j. This is because of the equation:

ad

_{H}(E

_{ij}) = (a

_{i}- a

_{j}) E

_{ij}

_{}

_{}(When i = j, E

_{ii}has trace 1 and is not in sl

_{3}(ℂ))

So, let us define the weight α

_{ij}: h -> ℂ by the formula α

_{ij}(H) = a

_{i}- a

_{j}. Then the above equation reads:

ad

_{H}(E

_{ij}) = α

_{ij}(H) E

_{ij}

_{}

To really get a picture of the weights of the adjoint representation, note that:

α

_{ij}= -α

_{ji}, α

_{ii}= 0, and α

_{ij}+ α

_{jk}= α

_{ik}.

_{}The above relations imply that all the weights of the adjoint representation can be expressed as a linear combination of two weights. Let us write:

α = α

_{12}, and β = α

_{23}.

_{13}= α + β, α

_{21}= -α, α

_{32}= -β, and α

_{31}= -α - β. To get a picture, let us plot these weights on the αβ-plane:

_{3}(ℂ)) = 8.

That's just the adjoint representation, but it turns out that all representations of sl

_{3}(ℂ) have pictures like this.

**The Eightfold Way**

*many*new particles. Such strongly interacting particles are called

**hadrons**- the Greek root

*hadros*means strong.

**strangeness**s(X), which was also an integer.

**spin-0 meson octet**(mesons are a type of hadron):

(Image by Laurascudder for Wikipedia.)

And here is the

**spin-1/2 baryon octet**. (Baryons are another type of hadron, which includes the proton and neutron. In fact, the proton and neutron are the particles n and p at the top):

(Image by Laurascudder for Wikipedia.)

As you can clearly see, both of these are pictures of the adjoint representation of sl

_{3}(ℂ)! This led the American physicist Murray Gell-Mann and the Israeli physicist Yuval Ne'emann to propose independently

**the eightfold way**hypothesis. The name comes from appearance of eight particles in the octets, and was Gell-Mann's allusion to the eightfold path to enlightenment in Buddhism. Here is the hypothesis:

**Hypothesis.**(The eightfold way)

*The hadrons are classified by representations of sl*

_{3}(ℂ).##### Representations of Lie groups

31 Janeiro 2022, 16:30 • John Huerta

Let G be a Lie group, and g = Lie(G) its Lie algebra. In the first half of this course, we studied representations for finite groups. We now wish to study them for Lie groups.

**Definition.** A
**representation **of G is a homomorphism π: G -> GL(V), where V is a finite-dimensional vector space over ℝ or ℂ.

**Example.** Every Lie group G has a god-given representation on its Lie algebra g, called the
**adjoint**
**representation** of G, and denoted Ad: G -> GL(g). If G is a matrix Lie group, then Ad is given by conjugation of matrices:

Ad
_{A}(X) = AXA
^{-1} , for A in G and X in g.

More generally, if G is any Lie group, Ad comes from differentiating the conjugation action of G on itself.

We can also study representations of Lie algebras.

**Definition.** Let g be a Lie algebra over 𝕂 = ℝ or ℂ. A
**representation** of g is a homomorphism of Lie algebras ∏: g -> gl(V), where V is a finite-dimensional vector space over 𝕂.

**Example.** Every Lie algebra g has a god-given representation on itself, called the
**adjoint representation **of g, and denoted ad: g -> gl(g). An element X in g acts on Y in g by bracketing: ad
_{X}(Y) = [X,Y].

**Exercise.** Show that ad is a representation of g for any Lie algebra g. You will need the Jacobi identity!

Now we want to ask, how are representations of a Lie group G related to those of its Lie algebra g = Lie(G)? To understand this, let us reach into our toolbox in differential geometry: given any smooth map f: M -> N between manifolds M and N, and a point p of M, we get a linear map between tangent spaces, f': T
_{p}M -> T
_{f(p)}N, the
**pushforward.**

Now apply this tool to a homomorphism of Lie groups, φ: G -> H. Let us take the pushforward at the identity:

φ': T
_{e}G -> T
_{e}H.

The second tangent space is at e because φ(e) = e - it's a group homomorphism! But we know these tangent spaces by another name; they're the Lie algebras of G and H, respectively. So we get a linear map:

φ': g -> h

**Fact.** The linear map φ' is a Lie algebra homomorphism.

Thus, from Lie group homomorphisms, we get Lie algebra homomorphisms by pushforward at e. Finally, let us apply this idea to representations. A representation of G is a homomorphism π: G -> GL(V). Taking the pushforward at e, we get a Lie algebra representation π': g -> gl(V).

**Key question**: can we go backwards, and get Lie group representations from Lie algebra representations?

It turns out the answer is yes, if we assume that G satisfies a topological condition: G must be "simply connected".

**Definition. **A topological space X is a
**simply connected** if

- for any two points x and y, there is a continuous curve connecting them: that is, there is a map c: [0,1] -> X such that c(0) = x and c(1) = y.
- if c
_{1}and c_{2}are two curves connecting x and y, c_{1}can be continously deformed into c_{2}through a family of curves that connect x and y: we say that c_{1}and c_{2 }are**homotopic.**

**Proposition.**If G is a simply-connected Lie group with Lie algebra g = Lie(G), then any finite-dimensional representation ∏: g -> gl(V) of the Lie algebra comes from a unique representation π: G -> GL(V) of the Lie group via pushforward: ∏ = π'.

This propostion is a big help! Representations of Lie algebras are essentially just linear algebra, so they are much easier to study that representations of Lie groups. Easiest of all is linear algebra over the complex numbers ℂ, because then we can always find eigenvalues and try to diagonalize matrices.

In light of this, we will focus on one family of examples. We will study the representations of the complex special linear group, SL

_{n}(ℂ).

**Fact.**The complex special linear groups SL

_{n}(ℂ) is simply connected for all n ≥ 1. Hence the complex, finite-dimensional representations of the Lie group SL

_{n}(ℂ) are the same as those of the Lie algebra sl

_{n}(ℂ).

SL

_{1}(ℂ) is the trivial group. So the first interesting example is SL

_{2}(ℂ), and its Lie algebra sl

_{2}(ℂ). It is this case that we will focus on for this lecture and most of the next.

In any case, the idea is always to try to diagonalize as many matrices as possible:

**Definition.**A

**Cartan subalgebra**h of sl

_{n}(ℂ) is a maximal abelian subalgebra such that the adjoint action ad

_{H}of any H in h can be diagonalized.

**Example.**Let h be the diagonal matrices in sl

_{n}(ℂ). This subalgebra is:

- abelian, because diagonal matrices commute;
- maximal, because any additional matrices would be off diagonal, and would fail to commute;
- ad
_{H }is diagonalizable: for H the diagonal matrix with entries a_{1}, a_{2}, ..., a_{n}, and E_{ij}the elementary matrix with 1 in the ijth entry and 0 elsewhere, we have ad_{H}(E_{ij}) = (a_{i}- a_{j}) E_{ij}. This shows that E_{ij }is an eigenvector of ad_{H}with eigenvalue a_{i}- a_{j}, and we can use this formula to find a basis of eigenvectors in sl_{n}(ℂ).

For sl

_{2}(ℂ), there's a Cartan subalgebra h = span{ H }, spanned by just one element:

H =

1 | 0 |

0 | -1 |

In fact, we have a standard basis for sl

_{2}(ℂ) including the element H:

1 | 0 |

0 | -1 |

F =

0 0

1 0

This basis satisfies the important relations:

[H, E] = 2E, [E,F] = H, [H, F] = -2F.

To help us analyze the representations of sl

_{2}(ℂ), we note the following theorem without proof:

**Theorem.**All complex, finite-dimensional representations of sl

_{n}(ℂ) are completely reducible.

So, any representation can be decomposed into a direct sum of irreducible representations (irreps). From now on, we will assume that the representation V is irreducible. The key result for understanding V is a bit deep:

**Deep theorem.**Given any finite-dimensional complex representation (V, ∏) of sl

_{2}(ℂ), then ∏

_{H}is diagonalizable.

This should be plausible, because H itself is a diagonal matrix, and we already know that ad

_{H}is diagonalizable. The miracle is that ∏

_{H}is diagonalizable for all ∏.

We use this diagonalizabity as follows: decompose the irrep V into eigenspaces:

V = ⊕

_{a}V

_{a}

_{}

where the direct sum is over all complex numbers a which are eigenvalues of ∏

_{H}, and each summand V

_{a}is the eigenspace for a. In other words, for all v in V

_{a}, we have Hv = av (really ∏

_{H}v = av, but I am going to suppress ∏).

We know how H acts - it acts by these eigenvalues. Next, we need to know how the other basis elements E and F act:

**Proposition.**If v is in V

_{a}, then Ev is in V

_{a+2}and Fv is in V

_{a-2}. (Really ∏

_{E}v and ∏

_{F}v, but I am going to suppress ∏.)

The proof of this proposition is so important, we call it the

**fundamental calculation**:

HEv = EHv + [H, E]v = aEv + 2Ev = (a + 2)Ev.

Similarly, HFv = (a - 2)Fv. This is what we wanted to check.

So, we have arrived at this picture of the representation V:

V = ... ⊕ V

_{a-2}⊕ V

_{a}⊕ V

_{a+2}⊕ ...

where:

- E acts by raising the eigenvalue by 2;
- F acts by lowering the eigenvalue by 2;
- H acts by multiplying by the eigenvalue.

##### The exponential map

27 Janeiro 2022, 12:30 • John Huerta

Let G be a Lie group with Lie algebra g = T
_{e}G. We want to understand the Lie algebra g better, and its relationship to G.

To help us, let us define:

**Definition.** A
*one-parameter subgroup* of G is a homomorphism φ: ℝ -> G.

Note that this is not actually a subgroup. It's a smooth map φ: ℝ -> G such that

- φ(0) = e;
- φ(s + t) = φ(s) φ(t) for all s and t in ℝ.

*homomorphism*is a smooth map φ: G -> H which is also a group homomorphism: φ(xy) = φ(x) φ(y).)

By differentiating at t = 0, every one-parameter subgroup gives us an element of the Lie algebra: X = φ'(0) is in g. Conversely, though it is not obvious, every element of g determines a unique one-parameter subgroup.

To see how this works, let us assume G is a matrix Lie group. That is, G is a closed subgroup of GL

_{n}(𝕂) for 𝕂 = ℝ or ℂ. Consider φ: ℝ -> G a one-parameter subgroup of G, with X = φ'(0) in g.

Note that for any integer N and real number t, we have:

φ(t) = φ(t/N)

^{N}.

For large N, the smooth function φ is well-approximated by its Taylor series:

φ(t/N) ≈ φ(0) + t/N φ'(0) + ...

(Note that we can add these terms because they are matrices.)

So approximately:

φ(t/N) ≈ I + t/N X

We substitute this into our formula:

φ(t) = φ(t/N)

^{N}≈ (I + t/N X)

^{N}

^{}

^{}I haven't proved it, but in the limit as N goes to infinity, this approximation becomes exact:

φ(t) = lim

_{N->∞}(I + t/N X)

^{N}

^{}

^{}Do you recognize the function on the right-hand side? It's the

*exponential function*of matrices:

φ(t) = exp(tX)

**Theorem.**For G a Lie group with Lie algebra g, there is a smooth map

exp: g -> G

called the

*exponential map*, satisfying the following properties:

- exp: g -> G is a local diffeomorphism at 0 in g.
- For any one-parameter subgroup φ: ℝ -> G, with X = φ'(0), we have φ(t) = exp(tX) for all t in ℝ.
- If G is a matrix Lie group, then exp is given by the usual power series for the exponential function: exp(X) = I + X + X
^{2}/2! + ... + X^{n}/n! + ... - If G is a matrix Lie group, then its Lie algebra is g = { X in gl
_{n}(𝕂) : exp(tX) is in G, for all t in ℝ}.

Property 4 gives us a way to compute the Lie algebras of matrix Lie groups, a tool we have wanted. Here is an example; more examples are in the lectures notes under Exercise 4.1.1!

**Example.**Show that sl

_{n}(ℝ), the n x n real traceless matrices, is the Lie algebra of SL

_{n}(ℝ). In addition, compute the dimension of the Lie group, dim(SL

_{n}(ℝ)).

**Solution.**Let us temporarily denote the Lie algebra by g = Lie(SL

_{n}(ℝ)). We want to show that g = sl

_{n}(ℝ). We will need a matrix identity:

det(exp(X)) = exp(tr(X)).

This holds for all n x n matrices with entries in ℝ or ℂ. By the fourth property in the theorem, we know that X is in g if and only if exp(tX) is in G for all t in ℝ. Since G = SL

_{n}(ℝ), this means that det(exp(tX)) = 1 for all t. By the identity, we have exp(t tr(X)) = 1 for all t. Differentiating in t and setting t = 0, we conclude tr(X) = 0. So, X is in sl

_{n}(ℝ). Conversely, if X is in sl

_{n}(ℝ), X is traceless, and the identity shows det(exp(tX)) = 1 for all t.

In sum, we conclude g = sl

_{n}(ℝ) as desired.

_{n}(ℝ)) = dim(sl

_{n}(ℝ)), and sl

_{n}(ℝ) is a vector space, so we just need to find a basis. As you can check, there is a basis for sl

_{n}(ℝ) consisting of n

^{2}- 1 matrices, so we have dim(SL

_{n}(ℝ)) = n

^{2}- 1.

##### Lie groups and Lie algebras

24 Janeiro 2022, 16:30 • John Huerta

We recalled the definition of Lie group, and gave abundant examples:

- (ℝ
^{n}, +) is an abelian Lie group. - The real general linear group GL
_{n}(ℝ) is, first of all, an open set in the set of n x n real matrices, Mat_{n}(ℝ). This makes it a manifold with one chart, and it is easy to check the group operations, given by matrix multiplication and matrix inversion, are both smooth. Hence, GL_{n}(ℝ) is a Lie group. - Similarly, the complex general linear group GL
_{n}(ℂ) is a Lie group. - (a) The real special linear group SL
_{n}(ℝ) is the subgroup of GL_{n}(ℝ) consisting of matrices with determinant 1. Because the determinant is a continous function, SL_{n}(ℝ) is a closed subgroup. Hence, SL_{n}(ℝ) is a Lie group by Cartan's theorem. Similarly, the complex special linear group SL_{n}(ℂ) is a Lie group.

- (b) The real orthogonal group O
_{n}(ℝ) is the subgroup of GL_{n}(ℝ) consisting of matrices A that satisfy A^{T}A = I. Because the map sending A to A^{T}A is continuous, this is a closed subgroup. Hence, O_{n}(ℝ) is a Lie group by Cartan's theorem. Similarly, the complex orthogonal group O_{n}(ℂ) is a Lie group. - (c) The Lorentz group O(1,n) is the subgroup of GL
_{n+1}(ℝ) consisting of matrices that satisfy A^{T}hA = h, where h is the (n+1) x (n+1) diagonal matrix with diagonal [-1, 1, 1, ..., 1]. - (d) The unitary group U
_{n}is the subgroup of GL_{n}(ℂ) of matrices satisfying A*A = I, where A* denotes the conjugate transpose of A. The special unitary group SU_{n}is the subgroup of GL_{n}(ℂ) satisfying A* A = I and det(A) = 1. - (e) The symplectic group Sp
_{n}(𝕂) is the subgroup of GL_{2n}(𝕂) of matrices satisfying A^{T}ΩA = Ω. Here, 𝕂 denotes ℝ or ℂ and Ω denotes the 2n x 2n that looks like this (each block is n x n):

0 -I I 0

- ℝ
^{n}with zero bracket, [v,w] = 0, is an abelian Lie algebra. - The real general linear algebra gl
_{n}(ℝ) is defined on the vector space Mat_{n}(ℝ), with Lie bracket given by the commutator: [X,Y] = XY - YX. - The complex general linear algebra gl
_{n}(ℂ) is defined on the vector space Mat_{n}(ℂ), with Lie bracket given by the commutator: [X,Y] = XY - YX. - (a) The real special linear algebra sl
_{n}(𝕂) is the Lie subalgebra of gl_{n}(𝕂) consisting of matrices of zero trace, for 𝕂 = ℝ or ℂ.

- (b) The real orthogonal algebra o
_{n}(𝕂) is the Lie subalgebra of gl_{n}(𝕂) consisting of matrices X that satisfy X^{T}+ X = 0. - (c) The Lorentz algebra o(1,n) is the Lie subalgebra of gl
_{n+1}(ℝ) consisting of matrices that satisfy X^{T}h + hX = 0, where h is the (n+1) x (n+1) diagonal matrix with diagonal [-1, 1, 1, ..., 1]. - (d) The unitary algebra u
_{n}is the Lie subalgebra of gl_{n}(ℂ) of matrices satisfying X*+ X = 0, where X* denotes the conjugate transpose of X. The special unitary algebra su_{n}is the Lie subalgebra of gl_{n}(ℂ) satisfying X* + X = 0 and tr(X) = 0. - (e) The symplectic algebra sp
_{n}(𝕂) is the Lie subalgebra of gl_{2n}(𝕂) of matrices satisfying X^{T}Ω + ΩX = 0, where Ω is the same matrix as above.

_{e}G. Naturally, for each of the Lie groups in the first list, its Lie algebra is the corresponding entry in the second list. So far, we only have enough differential geometry to check that T

_{e}G is a subspace of g, as we showed in the example where G = O(1,n), the Lorentz group, and g = o(1,n), the Lorentz algebra. Next time, we will see how to get equality.

##### Lie groups

20 Janeiro 2022, 12:30 • John Huerta

We completed our differential geometry toolkit with the Lie bracket of vector fields: given any two vector fields X and Y are a manifold M, the
**Lie bracket** [X,Y] on M is the operator on smooth functions defined by taking the commutator of X and Y:

**Prop**. [X,Y] is a vector field on M.

Using coordinates for the proof makes essential use of the fact that partial derivatives commute.

We then defined a Lie group: a

**Lie group**G is a group which is also a manifold such that the group operations are smooth. We stated Cartan's theorem on closed subgroups, and finished up by giving several examples of Lie groups:

- (ℝ
^{n}, +) is an abelian Lie group. - The real general linear group GL
_{n}(ℝ) is, first of all, an open set in the set of n x n real matrices, Mat_{n}(ℝ). This makes it a manifold with one chart, and it is easy to check the group operations, given by matrix multiplication and matrix inversion, are both smooth. Hence, GL_{n}(ℝ) is a Lie group. - Similarly, the complex general linear group GL
_{n}(ℂ) is a Lie group. - (a) The real special linear group SL
_{n}(ℝ) is the subgroup of GL_{n}(ℝ) consisting of matrices with determinant 1. Because the determinant is a continous function, SL_{n}(ℝ) is a closed subgroup. Hence, SL_{n}(ℝ) is a Lie group by Cartan's theorem. Similarly, the complex special linear group SL_{n}(ℂ) is a Lie group.

- (b) The real orthogonal group O
_{n}(ℝ) is the subgroup of GL_{n}(ℝ) consisting of matrices A that satisfy A^{T}A = I. Because the map sending A to A^{T}A is continuous, this is a closed subgroup. Hence, O_{n}(ℝ) is a Lie group by Cartan's theorem. Similarly, the complex orthogonal group O_{n}(ℂ) is a Lie group.