Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have to prove that to each of following classes of matrices can be given a manifold structure:

  1. symmetric (denoted with $\mathcal{S}$)
  2. upper triangular
  3. diagonal
  4. 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.

share|improve this question
1  
1), 3), and 4) are vector spaces, and 2) is an affine space; they're all isomorphic as manifolds to $\mathbb{R}^m$ for appropriate $m$. You need very few tools to see this. –  Qiaochu Yuan Nov 15 '11 at 18:58
1  
Indeed 2,3, and 4 are vector subspaces of $\mathfrak{gl}(n,\mathbb{R})$ –  Giuseppe Tortorella Nov 15 '11 at 18:59
1  
For 4), if the trace is 0 then $a_{nn}=-a_{11}-a_{22}-\cdots-a_{n-1,n-1}$. So you only need one chart mapping to $\mathbb{R}^{n^2-1}$ (just "forget" the $a_{nn}$ entry -- it's redundant). –  Bill Cook Nov 15 '11 at 19:08

2 Answers 2

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 $a_{ij}$:

  1. $a_{ij}-a_{ji}=0$ for $i>j$
  2. $a_{ij}=0$ for $i>j$
  3. $a_{ij}=0$ for $i\ne j$
  4. $a_{11}+\dots +a_{nn}=0$
share|improve this answer

I suggesst using this theorem from pafe 11 of https://people.maths.ox.ac.uk/hitchin/hitchinnotes/manifolds2012.pdf . Let me cite it

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. Than $F^{-1}[c] \subset \mathbb{R}^n$ has the structure of a $n$-differentiable manifold which is further more Hausdorff and with countable basis.

  • 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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.