Symmetric, upper triangular, diagonal and null-trace matrix spaces: are they manifolds? I have to prove that to each of following classes of matrices can be given a manifold structure:


*

*symmetric (denoted with $\mathcal{S}$)

*upper triangular

*diagonal 

*null trace.


I am interested in rather simply proofs that do not allow methods from the theory of Lie groups, instead tools like implicit functions theorem, basic results from basic topology, very basic results from manifold theory are allowed.



*

*About (1) 


I believe that since $\mathcal{S}$ is isomorphic to the upper right triangular matrices, we can consider the bijection
$$ M \in \mathcal{S} \mapsto (a_{11}\dots a_{1n},a_{22},\dots,a_{2n},a_{33},\dots a_{nn}) \in \mathbb{R}^{\frac{n(n+1)}{2}}$$
which defines a bijection. So we can trivially induce a topology on $\mathcal{S}$ from the standard topology over $\mathbb{R}^{\frac{n(n+1)}{2}}$, obtaining a single chart $C^\infty$ atlas (please correct if I'm wrong!!!).


*

*For (2) and (3) 


the reasoning is almost the same as in (1).


*

*About (4) 


I'd like to try with implicit function theorem, taking the space of $n \times n$ matrices ($n^2$-manifold) as starting point, considering the defining equation $f = \sum_i a_{ii} = 0$ and observing that the gradient of $f$ is not $0$ for each null trace matrix. So we can conclude that the space of null trace matrices is a closed submanifold of $M(n,\mathbb{R})$ of dimension $n^2-1$.
Is the preceding reasoning correct? Thanks.
 A: All four sets are linear subspaces of $\mathbb R^{n\times n}$, which can be inferred from their description by linear equations in terms of the matrix entries $a_{ij}$:


*

*$a_{ij}-a_{ji}=0$ for $i>j$

*$a_{ij}=0$ for $i>j$

*$a_{ij}=0$ for $i\ne j$

*$a_{11}+\dots +a_{nn}=0$

A: I suggest referencing the following theorem:
Theorem. Let $F:U \rightarrow \mathbb{R}^m$ be a $C^\infty$ function on the open set $U \subset \mathbb{R}^{n+m}$. Let $c \in \mathbb{R}^m$. Assume that for each $a \in F^{-1}[c]$ the total derivative
$$
DF_{a} :  \mathbb{R}^{n+m} \rightarrow \mathbb{R}^m
$$
is surjective. Then $F^{-1}[c]  \subset \mathbb{R}^n$ has the structure of a $n$-differentiable manifold which is furthermore Hausdorff and with countable basis. [1]

*

*The set of $n \times n$ matrices with zero trace is $n^2 -1$ differentiable manifold.

Let
$$
F: M_n(\mathbb{R}) \cong \mathbb{R}^{n^2} \rightarrow \mathbb{R}: \quad F(M)= TrM.
$$
$F$ is a linear map so it is $C^\infty$ and also it's total derivative coincides with $F$ in each point of $M_n(\mathbb{R})$. It is clear that $F^{-1}[0]$ is the set of $n \times n$ matrices with zero trace. Further more, to see that the derivative is surjective let $r$ be a real number. Define $M = \{m_{i,j} \}_{i,j=1}^n:$
$$
 m_{i,j}= 
        \begin{cases}
       r,& \{i,j\}=\{1,1\}
       \\
       0,&|i|+|j| >2       
         \end{cases},
$$
Then $F(M)=TrM = r$. So we have a $n^2-1$ differential manifold.

*

*The set of the symmetric $n \times n$ matrices  $S_n(\mathbb{R}) = \{ M \in M_n(\mathbb{R}) | M = M^T \}$ is $n(n+1)/2$ differentiable manifold.

Let
$$
F: M_n(\mathbb{R}) \cong \mathbb{R}^{n^2} \rightarrow A_n(\mathbb{R}) \cong \mathbb{R}^{n(n-1)/2}: \quad F(M)=\frac{1}{2}(M-M^T),
$$
where $A_n(\mathbb{R})$ denotes the linear space of all anti-symmetric matrices. Note that $F$ is the natural projection from $M_n(\mathbb{R})$ to $A_n(\mathbb{R})$ which follows from the direct sum
$$
M_n(\mathbb{R}) \cong S_n(\mathbb{R}) \oplus A_n(\mathbb{R}), \quad M_n(\mathbb{R}) \ni M = \frac{1}{2}(M+M^T)+\frac{1}{2}(M-M^T)
$$
Again, $F$ is a linear map so it is $C^\infty$ and also it's total derivative coincides with $F$ in each point of $M_n(\mathbb{R})$. It is clear that $F^{-1}[0]$ is the set $S_n$. Further more, to see that the derivative is surjective let $A=(a_{i,j})_{i,j=1}^n$ be an anti-symmetric matrix. Define $M = \{m_{i,j} \}_{i,j=1}^n:$
$$
 m_{i,j}= 
        \begin{cases}
       2a_{i,j}, & i\leq j 
       \\
       0, & i>j        
         \end{cases},
$$
then $F(M) = A$. So we have that $S_n$ is $n^2-n(n-1)/2 = n(n+1)/2$ differential manifold.

*

*For the space of Diagonal matrices we just use the fact that it is directly isomorphic to $\mathbb{R}^n$.


*For the space of upper-triangular matrices use projection onto the $n(n-1)/2$ dimensional subspace of matrices with 0 on and above the main diagonal.
[1] Page 11 of DIFFERENTIABLE MANIFOLDS, Course C3.1b 2012 by Nigel Hitchin
