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While reading manifold theory I stuck to this problem: $V$ be a vector space with $\dim V<\infty$ over $\mathbb{R}$ and $GL(V)$ be the group of all linear isomorphisms of $V$ into itself. A basis $e_1,\dots,e_n$ for $V$ induces a bijection $$GL_n(\mathbb{R}\longrightarrow GL(V),$$ $$[a^i_j]\mapsto (e_j\mapsto\sum_{i}a^i_je_i),$$ making $GL(V)$ into a $\mathcal{C}^{\infty}$ manifold, which we denote temporarily by $GL(V)_e$.

If $GL(V)_u$ is the manifold structure induced from the basis $u_1,\dots,u_n$ then how would you show they are diffeomorphic? Also I am not clearly understanding about the manifold structure of $GL(V)$, will be pleased for detail reply.

Would you give me the maximal atlas on $GL(V)$, coordinate charts, and $\mathcal{C}^{\infty}$ compatible maps? I guess If I some how get the those maps say $\phi_e$ and $\phi_u$ then $\phi_e o\phi_u$ would be a map by change of basis matrix $u$ to $e$ which wil be the diffeomorphism?

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Do you understand the manifold structure of $\textrm{GL}_n(\mathbb{R})$? The question doesn't really have much to do with $\textrm{GL}(V)$ – you can take $V = \mathbb{R}^n$ if you like. –  Zhen Lin Apr 25 '12 at 15:43
    
yes, I know that –  El Angel Exterminador Apr 25 '12 at 15:55

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up vote 3 down vote accepted

I think the point is that if you have two of these maps $\varphi, \psi\colon GL_n(\mathbb R) \to GL(V)$ determined by bases $\{e_1, \ldots, e_n\}$ and $\{f_1, \ldots, f_n\}$ then $\psi^{-1} \circ \varphi$ is a diffeomorphism: it is conjugation by the base change matrix $B = (b_{ij})$, where $e_j = \sum_{i = 1}^n b_{ij}f_i$. So the smooth structure does not depend on the basis that you choose.

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I'll write more later, if you like, but I have to hop on the train for a bit now. There's a chance that I messed up the indices, but it's more important to realize that there's nothing going on besides linear algebra. Do you understand how $GL_n(\mathbb R)$ gets its smooth structure as an open subset of $\operatorname{Mat}_{n \times n}(\mathbb R) \simeq \mathbb R^{n^2}$? –  Dylan Moreland Apr 25 '12 at 16:07
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Note also that the set of $n \times n$ matrices, and hence the open subset $GL_n(\mathbf R)$, has a global chart. Writing out a maximal atlas doesn't seem possible, but what the calculation I've done above shows you is that all of these global charts that you can give are contained in the same maximal atlas, so you can use whichever you like. –  Dylan Moreland Apr 25 '12 at 16:14
    
thank you Dylan –  El Angel Exterminador Apr 25 '12 at 18:07

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