Group homomorphism

In this section we define and discuss groups and group homomorphisms, including differentiable group homomorphisms, otherwise known as Lie group homomorphisms. 

It is often useful to think of relationships between various groups. To this end we define group homomorphisms and group isomorphisms. 

As an example, consider the detenninant. It is a standard result in linear algebra that if A and B are square matrices of the same size, then det( AB) = (detA)(detB). In other words, for each natural number n, the function det QL (C") C 0 is a group homomorphism. The kernel of the determinant is the set of matrices of determinant one. The kernel is itself a group, in this example and in general. See Exercise 4.4. A composition of... 

Definition 4.4 An injective group homomorphism 4 Gi G2 whose inverse is a group homomorphism from G2 to Gi is a group isomorphism. If there is a group isomorphism from a group G to another group G2, we say that the groups Gi anz/ G2 are isomorphic. 

Definition 4.6 Suppose Gi and G-. are Lie groups. Suppose 4 Gi —> G2 is a group homomorphism. If is differentiable, then T is a Lie group homomorphism. If is a also a group isomorphism and is differentiable then... 

There is an important surjective group homomorphism from 5 U (2) to 50(3). We will find the homomorphism useful in Section 6.6 for deriving the list of irreducible representations of 50(3) from the list of irreducible representations of 50(2). There is no a priori reason to expect such a homomorphism between two arbitrary groups, so the fact that 50(2) and 50(3) are related in this way is quite special. Here is the definition of 4> ... 

Hence is a group homomorphism. In the dehning formula for given in Equation 4.2, every matrix entry is a differentiable function of the real parameters gft(a), (a), and S( ). Because these parameters are differentiable functions on SU (2), the function is differentiable. Hence 4> is a Lie group homomorphism. 

Thus we have shown that any rotation around the z- or x-axis is in the image of the group homomorphism . Because any element of S O (3) can be written as a product of three such rotations (by Exercise 4.24). and because 4> is a group homomorphism, it follows that any element of 5 O (3) is in the image of . It remains only to show that 4> is two-to-one. Note first that... 

Any physical representation p must also be a Lie group homomorphism ... 

Exercise 4.4 Eix a natural number n and show that the set of n x n matrices of determinant one forms a group. Show more generally that the kernel of any group homomorphism is itself a group. Does the set of all matrices (of all finite sizes) of determinant one form a group under the usual matrix multiplication ... 

Exercise 4.9 (Used in Proposition 4.5) Suppose G and G are groups and 4 G G is a surjective group homomorphism. Suppose that the kernel of4> contains precisely n elements. Show that 4 is an n-to-one function, i.e., thatforany g e G the sef Yg contains precisely n elements. In particular, 4 is injective if and only = 1. 

What familiar condition on the quaternion u + xi + yj + "k w equivalent to requiring the corresponding matrix to be an element o/ 5(9(1, 3) Use this calculation to define a group homomorphism from the set of quaternions satisfying that condition to SO (A). 

Exercise 4.26 Show that there is an injective group homomorphism from 517(2) to 5(9(4). In other words, show that there is a subgroup of SO (A) that is isomorphic to SU(T). (Hint use quaternions.) Is this homomorphism surjective ... 

Exercise 4.34 (SU (2) and the unit quaternions) Recall the functions f, fj and f)i,from Exercise 2.8. Show that the restrictions off, f and fk to the unit circle T are group homomorphisms whose range lies in the unit quaternions. Call their images T, Tj and Tk, respectively. Write the images ofTj, Tj azjJ Tk under the homomorphism explicitly a5 3 x 3 matrices. 

In this section we show how to use group homomorphisms to construct a representation of one group from a representation of another group. 

Proof. First we show that (G, V, p o F) is a representation by checking the criteria given in Definition 4.7. We know by hypothesis that G is a group and V is a vector space. Because both p and 4 are group homomorphisms, it follows from Proposition 4.3 that p o group homomorphism from G to QC (V). Hence p o 4 is a representation. 

Consider for example the inclusion map i of a subgroup G of a group G, defined in Exercise 4.37. By that exercise, the inclusion map is a group homomorphism. Note that for any representation p of G, the pullback representation p o z is just the restriction of p to the subgroup G. 

It is not usually possible to push a representation forward, i.e., to use a representation on the domain of a group homomorphism to obtain a representation on the image. See Exercise 5.4. However, in certain circumstances a pushforward representation can be defined. See Figure 5.3. 

In Section 6.6 we will both push representations forward and pull them back along the two-to-one group homomorphism T 5(7(2) 50(3) intro-... 

Exercise 5.4 Find a representation p and a group homomorphism 4> such that p cannot be pushed forward via 4. ... 

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