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In a problem I'm currently tackling (not related to the question) the map $f : S^2 \times \mathbb{R}^3 \to \mathbb{R}^3$ is defined as

$$ (d,v) \to \langle v,d\rangle d = dd^T v $$

($S^2$ is the unit sphere in $\mathbb{R}^3$). As exercise I was trying to compute the tangent space of the manifold

$$ M = \left\{P \in \mathbb{R}^{3 \times 3} : P^2=P \right\} $$

as $dd^T \in M$ I thought this was a good exercise. My attempt was based on few examples I saw in chapter 15 and 16 of Tu's Introduction to Manifolds.

I'm going to define a curve $c : ]-\epsilon,+\epsilon[ \to M$, $\epsilon > 0$ such that $c(0) = P \in M$ and $c'(0) = X_P \in T_P(M)$. I'm also observing that if I define $f : \mathbb{R}^{3\times 3} \to \mathbb{R}^{3 \times 3}$

$$ P \to P^2 - P $$

then $M = f^{-1}(0)$ and it's a regular level set. I observe that $$ f_{*,P} X_P = \frac{d}{dt}(f \circ c)(0) = \frac{d}{dt}(c(t)c(t) - c(t))(0)= (c'(t)c(t) + c(t)c'(t) - c'(t))(0) = X_P P + P X_P - X_P $$

and because of the constraint on our manifold it must be

$$ X_P P + (P - I)X_P = 0 $$

Therefore I conclude

$$ T_P(M) = \left\{ X_P \in \mathbb{R}^{3 \times 3} : X_P P + (P - I)X_P = 0 \right\} $$

I have two questions then:

  1. Is my derivation correct?
  2. If yes, is there a nice geometric interpretation maybe I can spot here?
user8469759
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  • Note that $M$ is disconnected and the components are not of the same dimension. It also requires checking that the components are manifolds. – hunter Jul 14 '20 at 16:13
  • Hmmm, didn't think of that. How did you show it's disconnected? – user8469759 Jul 14 '20 at 16:16
  • (Noticed something was funny by plugging in $I$ and $0$ as a sanity check to your formula gives zero-dimensional tangent spaces, which makes sense as these are isolated points.) The map which takes $P$ to the coefficients of the characteristic polynomial of $P$ is a continuous map to $\mathbb{R}^4$, and has a disconnected image since there are only 4 possible characteristic polynomials. – hunter Jul 14 '20 at 16:21
  • Is my derivation wrong then? Or is this a geometric interpretation? – user8469759 Jul 14 '20 at 16:27
  • Since any matrix is conjugate over $\mathbb{R}$ to its rational canonical form, and all these matrices are diagonalizable (minimal poly has distinct roots) the space is going to be the union of quotients of $GL_3(\mathbb{R})$ by the centralizers of the four diagonal matrices whose diagonal entries are 0's and 1's. (order doesn't matter). – hunter Jul 14 '20 at 16:27
  • (haven't checked your derivation yet, it could totally be right; you still need to check that this is a manifold though since the calculus might be meaningless at a bad point.) – hunter Jul 14 '20 at 16:28
  • @hunter, isn't map the takes $P$ and returns the coefficients of the characteristic polynomial a map to $\mathbb{R}^3$ and not $\mathbb{R}^4$? – user8469759 Jul 16 '20 at 18:45

1 Answers1

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Your derivation is correct.

An informal way of thinking about it: you are solving for $A$ such that $(P + \epsilon X)^2 = (P+\epsilon X)$, where you can treat $\epsilon^2$ as zero. This gives the same equation for the tangent space at $P$.

As noted in the comments, the manifold actually has four components of different dimensions (two of which are just points). Your appeal to the regular value theorem is correct and proves that each component is a manifold (there are two zero-dimensional components and two four-dimensional components).

hunter
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  • It's still not clear to me why exactly is not connected. Could you clarify this as well? Maybe this will give me an hint on how to prove this is a manifold. – user8469759 Jul 14 '20 at 18:19
  • @user8469759 for example, the determinant map is a continuous map from this set to $\mathbb{R}$ and it lands on ${0, 1}$. So this set can't be connected. To see there are at least 4 components, look at the determinant and the trace. To see that these are all the components: every matrix in your space is conjugate to a diagonal matrix with only $1$s and $0$s on the diagonal, and each conjugacy class is connected since it's an image of $\text{GL}_3(\mathbb{R})^+$. – hunter Jul 14 '20 at 18:31