91影视

Let \(A\) be the subgroup of \(G\) consisting of all matrices $$s(a)=\left(\begin{array}{cc} a & 0 \\ 0 & a^{-1} \end{array}\right) \quad \text { with } a>0 .$$ (a) Show that the map \(a \mapsto s(a)\) is a homomorphism of \(\mathbf{R}^{*}\) into \(G\). Since this homomorphism is obviously injective, this homomorphism gives an imbedding of \(\mathbf{R}^{+}\) into \(G\). (b) Let \(U\) be the subgroup of \(G\) consisting of all clements $$u(x)=\left(\begin{array}{ll} 1 & x \\ 0 & 1 \end{array}\right) \quad \text { with } x \in \mathbf{R}$$ Thus \(u \mapsto u(x)\) gives an imbedding of \(\mathbf{R}\) into \(G\). Then \(U A\) is a subset of \(G\). Show that \(U A\) is a subgroup. How does it differ from the Borel subgroup of \(G\) ? Show that \(U\) is normal in \(U A\). (c) Show that the map $$U A \rightarrow \mathbf{H}$$ given by $$\beta \mapsto \beta(i)$$ gives a bijection of \(U A\) onto \(\mathrm{H}\). (d) Show that every element of \(\mathrm{SL}_{2}(\mathbf{R})\) admits a unique expression as a product $$u(x) s(a) r(\theta),$$ so in particular, \(G=U A K\).

Step by step solution

01

Verify the map is a homomorphism

We need to show that for any real numbers a, b the map satisfies: \(s(a\cdot b)=s(a)s(b)\) We calculate both sides: \(s(a\cdot b) = \begin{pmatrix}ab & 0 \\ 0 & (ab)^{-1}\end{pmatrix}\) \(s(a)s(b) = \begin{pmatrix}a & 0 \\ 0 & a^{-1}\end{pmatrix} \begin{pmatrix}b & 0 \\ 0 & b^{-1}\end{pmatrix} = \begin{pmatrix}ab & 0 \\ 0 & (ab)^{-1}\end{pmatrix}\) Since both sides are equal, the map is a homomorphism.

02

Show injectivity

We need to show that the map has a unique corresponding value for every distinct a in 鈩濃伜: Suppose \(s(a_1)=s(a_2)\) for some \(a_1, a_2 \in \mathbf{R^{+}}\). Then: \(\begin{pmatrix}a_1 & 0 \\ 0 & a_1^{-1}\end{pmatrix} = \begin{pmatrix}a_2 & 0 \\ 0 & a_2^{-1}\end{pmatrix}\) It is clear that \(a_1 = a_2\), implying injectivity. #b. Subgroup and Normality#

03

Show UA is a subgroup

To show that UA is a subgroup, we need to show that it satisfies closure, identity, and inverse properties. 1. Closure: If \(u(x_1)s(a_1), u(x_2)s(a_2) \in UA\), then their product is also in UA: \(u(x_1)s(a_1)u(x_2)s(a_2) = u(x_1)(u(x_2)s(a_2))s(a_1) = u(x_1)u(x_2)s(a_2a_1) = u(x) s(a) \in UA\) 2. Identity: The identity matrix is also in UA: \(u(0) s(1) = I\) 3. Inverse: For any element \(u(x) s(a) \in UA\), its inverse is: \((u(x) s(a))^{-1} = s(a)^{-1} u(x)^{-1} = u(-x) s(a^{-1}) \in UA\) Therefore, UA is a subgroup. The Borel subgroup is slightly different because it contains diagonal matrices with non-equal diagonal entries, while A does not.

04

Show U is normal in UA

To show that U is normal in UA, we need to show that for any element \(u(x) \in U\), \(u(x)^{-1} \cdot u_a \cdot u(x) \in U\), where \(u_a = u(y)s(a) \in UA\). \(u(x)^{-1} \cdot u(y) s(a) \cdot u(x) = u(-x)u(y) s(a) u(x) = u(-x+y) s(a) u(x)\) As \(s(a)u(x)u(a)^{-1} = u(a)u(a)^{-1} = I\), the product is in U. Thus, U is normal in UA. #c. Bijection#

05

Verify bijection

The given map is: \(UA \rightarrow H\), \(\beta \mapsto \beta(i)\). By definition, a bijection between two sets must be both injective and surjective. Injectivity: Suppose the map is not injective. Then, there exist \(\beta_1, \beta_2 \in UA\) such that: \(\beta_1(i) = \beta_2(i)\) However, this contradicts the uniqueness of representation given in (d). So the map is injective. Surjectivity: Since UA includes all elements of the form \(u(x)s(a)\), and the map \(\beta \mapsto \beta(i)\) produces elements in H which can be represented by any \(u(x)s(a)\), we have covered all elements of H. Hence, the map is surjective. Thus, the given map is a bijection. #d. Unique expression and G=UAK#

06

Show unique expression

Any element in SL2(鈩�) can be decomposed as: \(\begin{pmatrix}a & b \\ c & d\end{pmatrix} = u(x) s(a) r(\theta)\) Where \(a > 0\), and \(cos(\theta) = \frac{a + d}{2a}\), \(sin(\theta) = \frac{b - c}{2a}\). Since the right side is a product of three factors, each containing different parameters, each element in SL2(鈩�) has a unique expression.

07

Show G=UAK

As mentioned earlier, both A and U are subgroups of G. Now, let K be a subgroup of SL2(鈩�) containing all elements of the form r(胃). Since every element of SL2(鈩�) admits a unique expression as a product u(x)s(a)r(胃), it is clear that G is equal to the product of U, A, and K, i.e., G=UAK. This completes the exercise.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Homomorphism

A homomorphism is a type of function between two algebraic structures, like groups or rings, that preserves their operations. Imagine a translator between two different languages. The key for a homomorphism is that it translates expressions in such a way that the structure and meaning remain consistent in both settings.

For example, consider the map \( s : \mathbf{R}^{*} \rightarrow G \) given by \( a \mapsto s(a) \), where \( s(a) = \begin{pmatrix} a & 0 \ 0 & a^{-1} \end{pmatrix} \). This map needs to maintain the same operation: if you multiply \( a \) and \( b \) in \( \mathbf{R}^{*} \), their image under \( s \), which is \( s(a)s(b) \), should be the same as \( s(a \cdot b) \). Since both expressions give us \( \begin{pmatrix} ab & 0 \ 0 & (ab)^{-1} \end{pmatrix} \), our map is indeed a homomorphism. This property is essential when embedding structures like \( \mathbf{R}^{+} \) into another group \( G \).

Subgroup

A subgroup is a smaller group contained within a larger group that also satisfies the group's operations. Think of it like a club within a bigger community that has its own rules but still abides by the general community rules. For a subset of a group to be a subgroup, it must include the identity element, be closed under the group operation, and contain inverses for all its elements.

Let's take the example from the exercise. We want to show that \( UA \) is a subgroup of \( G \). We check:

• **Closure:** This ensures the product of any two elements in the subgroup is still within the subgroup. So for \( u(x_1) s(a_1) \) and \( u(x_2) s(a_2) \), their product is \( u(x_1 + x_2) s(a_1 a_2) \), and it must also lie in \( UA \).
• **Identity:** The identity element of a group doesn't change other elements when used in the group operation. For \( UA \), this is \( u(0) s(1) = I \), the identity matrix.
• **Inverse:** For every element \( u(x) s(a) \), the inverse, such as \( u(-x) s(a^{-1}) \), must also be in \( UA \).

These checks affirm that \( UA \) is indeed a subgroup.

Injectivity

Injectivity in a map means every element from the original set maps to a distinct element in the target set. It's about a one-to-one relationship, where no two inputs have the same output.

In the context of our exercise, we examine the map \( a \mapsto s(a) \) to determine if it is injective. Suppose we have two values, \( a_1 \) and \( a_2 \) in \( \mathbf{R}^{+} \), such that their images are the same: \( s(a_1) = s(a_2) \). This implies:

\[ \begin{pmatrix} a_1 & 0 \ 0 & a_1^{-1} \end{pmatrix} = \begin{pmatrix} a_2 & 0 \ 0 & a_2^{-1} \end{pmatrix} \]

From the equality of the matrices, it is clear that \( a_1 = a_2 \). Therefore, each input maps uniquely to an output, which confirms that the homomorphism is injective. This property is pivotal in ensuring the structure of the subgroup \( UA \) is maintained appropriately during the mapping.

Normal Subgroup

A normal subgroup is a special type of subgroup that remains unchanged when you "conjugate" it by other elements from the larger group. It's like a group within a club that remains the same no matter how you shuffle the club members around. Normal subgroups are essential in the concept of group symmetry and the construction of quotient groups.

Considering our subgroup \( U \) within the larger subgroup \( UA \), we want to verify if \( U \) is normal in \( UA \). This requires:

• For any element \( u(x) \) in \( U \) and any element \( \ u(y) s(a) \) in \( UA \), the element \( u(x)^{-1} u(y) s(a) u(x)\) must also be in \( U \).

Upon simplifying:

\[ u(x)^{-1} u(y) s(a) u(x) = u(-x + y) s(a) u(x) \]

Since \( s(a) u(x) s(a)^{-1} \) remains in \( U \), the entire expression satisfies being in \( U \) again. Thus, \( U \) enjoys the property of being normal, meaning that this subgroup behaves well under conjugation with elements from \( UA \). This property is crucial in analyzing the symmetry and more extensive structure formed by \( UA \).