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How would I start off proving that $S= $(the set of symmetric $n\times n$ matrices) is a manifold. I tried using the definition directly by saying $M_n =$ the space of all $n\times n$ matrices For every $A\in M_n$ there exists open sets $U=V=M_n$ and a bijection $F: U\to V$ by $F(A)= A-A^T$ Therefore we have $F(U \cap S) = F(S)$ since $S$ is a subset of $M_n=\{0\} \cap M_n$ this is where I get stuck. Also, I know that the set of all symmetric $n\times n$ matrices is $(n^2 + n)/2$, therefore that is the dimension of the manifold

Definition: A set $M$ (subset of $\Bbb{R}^n$) is a $k$-dimensional manifold if for every $x\in M$ there exists open sets $U$, $V$ and a bijection $h:U\to V$ with $x\in U$ and $H(U \cap M) = V \cap (\Bbb{R}^k \times \{c^{k+1},\ldots ,c^n\})$ for all $c$'s constants

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    The set is a linear subspace...2010-11-03
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    How would I use that?2010-11-03
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    Every vector space has a basis, so there's a linear isomorphism with $\mathbb R^n$ where $n$ is the dimension of the space. This is your chart (you only need one).2010-11-03
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    What definition of a manifold are you using? You might be using a particular one where the above reasoning will require some tweaking.2010-11-03
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    Chart? I was under the impression that Mn is a subset of R^(n^2)2010-11-03
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    Technically your definition of "manifold" would be called a "submanifold of Euclidean space". "Manifold" in the common math world usually means "abstract manifold". To construct your map $H$ (which I call a chart), $U$ and $V$ will be equal to $n\times n$ matrices. Choose a basis for the space of matrices which includes a basis for your subspace, and represent matrices with respect to that basis. That's the idea.2010-11-03
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    I'm sorry... I took this Differential Geometry class before I took any linear algebra class since it wasn't a prerequisite so I'm just a little confused when you say I can choose my own basis2010-11-03
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    Should I scrap the proof I have? or am I in the right direction?2010-11-03
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    Oh, I'm afraid you're in for a pretty rough ride learning differential geometry without having linear algebra as background. I'd like to suggest you talk to your prof about what you're getting into.2010-11-03
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    I don't mean to imply it's impossible to do the task you've got set-out in front of you. Only, it will be much harder to grasp the essential ideas as it will take more effort to unwind the concepts into terms you're familiar with.2010-11-03

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