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

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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.

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?)

  1. 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.

2 Answers 2

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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.

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    So the first answer is that the polynomial is smooth. – 2017-01-02
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    Yes, and rational fractions are smooth as well for the inverse. – 2017-01-02
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    But I am still curious that the definition of Lie group only require such two maps be smooth? not for other functions like square, sine function...? Is this because these are the operations of group? – 2017-01-02
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    A Lie group is a manifold AND a group. For example, $\sin$ is not a group morphism (and even not well defined : how would you define $\sin$ for $x \in \mathbb R^3$ ?) So the only important thing is that the group morphisms is smooth. In some sense, theory of manifold is approximating everything with linear map : since we have a group we want to approximate multiplication and inverse by linear map. And this is a really successful approach. – 2017-01-02
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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.