NBHM 2020 PART C Question 31 Solution (continuous images of $G = GL_2(\Bbb R)$)

Let $G = GL_2(\Bbb R)$ be the set of all $2 \times 2$ invertible real matrices. Elements of $G$ can be identified with vectors in $\Bbb R^4$ and this makes $G$ a metric space. Which of the following sets can be obtained as the image of a continuous onto function from $G$? 
1. The real line. 
2. $A = \{(x,\frac{1}{x}) : x \in \Bbb R \text{ and } x \ne 0\}$.
3. $\Bbb R^2 \backslash A$. 
4. $\{(x,y) \in \Bbb R^2 : |x|^2 + |y^2| \le 1\}$
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Solution:
Option 1:(True) Consider the determinant map $\text{det }: GL_2(\Bbb R) \to \Bbb R^*$ where $\Bbb R^{*} = \Bbb R \backslash \{0\}$ is a group under multiplication. Now $$\text{det}(AB) = \text{det}(A)\text{det}(B)$$ Therefore this defines a homomorphism. We have determinant is a continuous map, because it is a polynomial in the matrix entries.
Let $\alpha \in \Bbb R^*$ then the matrix $\begin{bmatrix}\alpha & 0 \\ 0 & 1\end{bmatrix} \in GL_2(\Bbb R)$ has determinant $\alpha$. Therefore determinant is a surjective map. Now, the function $\phi: \Bbb R^* \to \Bbb R$ defined by $\phi(x) = |\text{log}(x)|$ is a continuous surjective map. Hence the composition $\phi \circ \text{det} : GL_2(\Bbb R) \to \Bbb R$ is the required continuous surjection. Note that this map is also a composition of group homomorphisms and hence is a group homomorphism. Therefor there exists a continuous surjective group homomorphism between $GL_2(\Bbb R)$ to $\Bbb R$. 
Option 2:(True) Consider the map $\phi : GL_2(\Bbb R) \to A = \{(x,\frac{1}{x}: x \in \Bbb R)\}$ defined by $\phi(A) = (\text{det}(A),\frac{1}{\text{det}(A)})$. We have seen that $\text{det}$ is a continuous map. Since $\text{det}(A) \ne 0$ for any $A \in GL_2(R)$, the function $A$ goes to $\frac{1}{\text{det}(A)}$ is also continuous. This shows that, the function $\phi$ is continuous since each of its coordinate map is continuous. Let $(\alpha,\frac{1}{\alpha}) \in A$ then the matrix $M$ defined by $\begin{bmatrix}\alpha & 0 \\ 0 & 1\end{bmatrix} \in GL_2(\Bbb R)$ has determinant $\alpha$ and $\phi (M) = (\alpha,\frac{1}{\alpha})$. Therefore $\phi$ is surjective.
option(3): 
Result: Let $X$ and $Y$ be two metric spaces such that $X$ has $m$ connected components and $Y$ has $n$ connected components with $m < n$. Then there cannot be a surjective continuous map from $X$ to $Y$.
Proof: Let $f$ be a continuous function from $X$ to $Y$ and $X_1,X_2,\dots,X_m$ be the connected components of $X$. We can restrict $f$ to each of these connected component and the resulting map is still continuous. Also, the images $f(X_i)$ is connected for each $i$. Let $Y_i$ be the connected component of $Y$ such that $f(X_i) \subseteq Y_i$  (Note that, different connected components $X_i$ could have mapped to the same connected component $Y_i$). This shows that $f(X) \subseteq \cup_{i=1}^m Y_i$. This means that $f(X)$ doesn't intersect remaining $n-m$ connected components of $Y$. Therefore $Y$ is not surjective.
 $GL_2(\Bbb R)$ is disconnected and has $2$ connected components. The set $\Bbb R^2 \backslash A$ has $3$ connected components. Now, the above result shows that there cannot be continuous surjection between these sets.
option 4:(True)
Consider the function $f$ from $GL_2(\Bbb R)$ to $\Bbb C$ given by $M = \begin{bmatrix}x & y \\ z & a\end{bmatrix}$ goes to $\frac{y^2}{x^2+ y^2}e^{i a}$. Then clearly $f$ is a continuous surjection from $GL_2(\Bbb R)$ to $\{(x,y) \in \Bbb R^2 : |x|^2 + |y^2| \le 1\}$.
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