The set $2\times 2$ invertible matrices forms a Lie group

Question

I am a super junior student in this field. The following comes from the wiki:
https://en.wikipedia.org/wiki/Lie_group

The following is what I know:

1. This is a four dimensional group since we can vectorize a $2\times 2$ matrix. So the set of $2\times 2$ matrices is isomorphic to $\mathbb{R}^4$.
2. It is noncompact because there is no limit on each entry; each entry can grow up to infinity.
3. It has two connected components since there are matrices with $0$ determinant.

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My question is the following:

1. To be a Lie group, the group operations of multiplication and inversion are smooth map, i.e., they are $C^{\infty}$-function. For example, $\mu(x,y) = xy$. How to understand this is a $C^{\infty}$-function? Just take derivative with respect to $x$ and $y$? I have no idea what does it mean exactly; hope for a detailed explanation.

(I roughly know $C^{\infty}$-manifold means the all transition functions are $C^{\infty}$-differentiable. But in the definition of Lie group we only consider two transition functions, i.e., multiplication and inverse?)

2. How to argue "GL$(2,\mathbb{R})$ is NOT path connected (but still connected)" is false?

Please open a door for me in these big topics.

Answer 1

1 : First it is a manifold because it is an open subset of $\mathbb R^4$. Coordinates are $x_{11}, x_{12}, x_{21},x_{22} $. For example, multiplication of matrix is simply the application $( x_{11}, x_{12}, x_{21},x_{22}), (y_{11}, y_{12}, y_{21},y_{22}) \mapsto (x_{11}y_{11} + x_{12}y_{21}, \dots ) \in \mathbb R^4$.Every component is a polynomial : in particular the map is $\mathbb C^{\infty}$. Applyint Cramer's rule will give you the $C^{\infty}$ for inverse.

2 : Think about the determinant.

Answer 2

re: $GL_n\Bbb{R}$ being compact. Topologically, $GL_n\Bbb{R}$ is an open subset of $\Bbb{R}^4$ (this is because $\det$ is continuous, so the preimage $S$ of this function at $0$ is closed, and so $GL_n\Bbb{R} = \Bbb{R}^4-S$ is open). It should then be clear that $GL_n\Bbb{R}$ isn't closed in $\Bbb{R}^4$, and so it isn't compact.

Matrix multiplication is smooth, since each component of the product is a polynomial of the original $8$ variables. Polynomials are smooth, and so multiplication is smooth.

Inversion is smooth since the components of the inverse of a matrix are rational functions of the matrix's components. This is evident by the formula $$A^{-1} = \frac{1}{\det A}\mathrm{adj}(A)$$ and by the fact that the determinant is a polynomial in terms of the entries of a matrix.

Finally, as you noted, $GL_n\Bbb{R}$ is not connected, and thus it is not path connected. To show that $GL_n\Bbb{R}$ is not path connected, let $\gamma:[0,1]\to GL_n\Bbb{R}$ be a path from $A$ to $B$, such that $\det A > 0, \det B < 0$. $\det(\gamma)$ is a function from $\Bbb{R}$ to $\Bbb{R}$, and applying the intermediate value theorem tells us that there exists $t\in (0,1)$ such that $\det(\gamma(t)) = 0$, but this is impossible. Thus, $GL_n\Bbb{R}$ is not path connected.