For the proof of Cohn-Vossen's rigidity theorem I need to prove the next lemma (can be found in Montiel-Ros's Curves and Surfaces page 218):

If $\Phi$ and $\Psi$ are two definite self-adjoint endomorphisms of a Euclidean vector plane and $det \Phi = det \Psi$. Then, $det (\Phi + \Psi) \leq 0$ and that equality occurs if and only if $\Phi = - \Psi$.

The proof says the following:

If $det(\Phi +\Psi) > 0$, the endomorphism $\Phi+\Psi$ would be definite, say, positive definite. Now, we take $\{ e_1, e_2 \}$ a basis of the plane diagonalizing $\Phi$, one has $$ \langle \Phi(e_i), e_i \rangle + \langle \Psi(e_i), e_i \rangle > 0, \;\;\;\; i = 1,2.$$ Consequently, $$ det \Phi = \langle \Phi(e_1), e_1 \rangle \langle \Phi(e_2), e_2 \rangle \stackrel{(1)}{>} \langle \Psi(e_1), e_1 \rangle \langle \Psi(e_2), e_2 \rangle \geq \langle \Psi(e_1), e_1 \rangle \langle \Psi(e_2), e_2 \rangle - \langle \Psi(e_1), e_2 \rangle^2 \stackrel{(2)}{=} det \Psi $$ which gives a contradiction to our hypothesis. Therefore, $det( \Phi + \Psi ) \leq 0$.

Here, I don't understand inequality (1) and equality (2).

Moreover, the lemma says the following:

In the above inequality, equality occurs if and only if $\Phi = - \Psi$.

The proof says the following:

Following the reasoning above, if equality holds, there wouldbe at least a non-null vector in the kernel of $\Phi + \Psi$. Let $\{u_1, u_2 \}$ be a basis diagonalizing $\Phi + \Psi$, that is, such that $$ \Phi(u_1) + \Psi(u_1) = 0 \;\;\;\; and \;\;\;\; \Phi(u_2) + \Psi(u_2) = \lambda u_2, \;\;\; \lambda \in \mathbb{R}. $$ From the first equality we deduce that $$ \langle \Phi(u_1), u_1 \rangle = - \langle \Psi(u_1), u_1 \rangle \;\;\;\; and \;\;\;\; \langle \Phi(u_1), u_2 \rangle = - \langle \Psi(u_1), u_2 \rangle, $$ which together with the facts that $det \Phi = det \Psi$ and that $\Phi$ and $\Psi$ are definite, gives the equality $$ \langle \Phi(u_2), u_2 \rangle \stackrel{(3)}{=} - \langle \Psi(u_2), u_2 \rangle, $$ implying $\lambda = 0$. Thus, in this case, $\Phi = - \Psi$.

Of this latest part I don't understand the equality (3).

Could you show me why these equations hold?

  • $\begingroup$ @TedShifrin great! I could figure out (2) with your comment! $\endgroup$
    – EvaMGG
    Oct 5, 2018 at 21:42
  • $\begingroup$ In addition to Ted's comment, notice that the lemma is trivially false if $\Phi$ and $\Psi$ are both the identity. $\endgroup$
    – user7530
    Oct 5, 2018 at 22:59
  • $\begingroup$ for a proof of the theorem you can look at page 87 : math.brown.edu/~deigen/chern.pdf $\endgroup$ Oct 7, 2018 at 17:06

2 Answers 2


Once I fixed a crucial typo, some things became clearer. Since we're assuming $\Phi+\Psi$ is positive definite, we have $\langle (\Phi+\Psi)e_i,e_i\rangle > 0$ for $i=1,2$, so $\langle\Phi(e_i),e_i\rangle > -\langle\Psi(e_i),e_i\rangle$ for $i=1,2$. "Therefore," so to speak, $$\langle\Phi(e_1)e_1\rangle\langle\Phi(e_2),e_2\rangle>\langle\Psi(e_1),e_1\rangle\langle\Psi(e_2),e_2\rangle,$$ establishing inequality (1). This is the proof the authors intended, but of course it's wrong unless we are assuming $\Psi$ is negative definite here, so that both right-hand sides are positive and we can multiply the inequalities.

Equality (2) is just the computation of $\det\Psi$ using a matrix representation with respect to the basis $\{e_1,e_2\}$.

Now that we've narrowed things down to assuming that $\Phi$ is positive definite and $\Psi$ is negative definite, let's look at (3). This is also following from a determinant computation: \begin{align*} \det\Phi&=\langle \Phi(u_1),u_1\rangle \langle \Phi(u_2),u_2\rangle - \langle \Phi(u_1),u_2\rangle^2\\ \det\Psi &= \langle \Psi(u_1),u_1\rangle\langle \Psi(u_2),u_2\rangle -\langle \Psi(u_1),u_2\rangle^2. \end{align*} Since $\det\Phi=\det\Psi$, substituting the first two equalities, we get $$\langle \Phi(u_1),u_1\rangle \langle \Phi(u_2),u_2\rangle - \langle \Phi(u_1),u_2\rangle^2 = -\langle \Psi(u_1),u_1\rangle \langle \Phi(u_2),u_2\rangle - \langle \Psi(u_1),u_2\rangle^2 = \langle \Psi(u_1),u_1\rangle\langle \Psi(u_2),u_2\rangle -\langle \Psi(u_1),u_2\rangle^2,$$ and so $\langle\Psi(u_2),u_2\rangle = -\langle \Phi(u_2),u_2\rangle$, as they claimed.

Let me reiterate that the lemma is false as stated. If we assume both $\Phi$ and $\Psi$ are positive definite, then we in fact should conclude that $\det(\Phi-\Psi)\le 0$ with equality holding iff $\Phi=\Psi$. If we assume (as apparently these authors meant to) that $\Phi$ is positive definite and $\Psi$ is negative definite, then we conclude that $\det(\Phi+\Psi)\le 0$ with equality holding iff $\Phi=-\Psi$.

  • $\begingroup$ @EvaMGG: Sorry about my own typos. Generally, rather than editing, it's better just to put a comment to get it clarified. (I mistakenly rejected the edit because I couldn't believe I had it wrong!! :P) ... Anyhow, I hope the situation is now clear. P.S. I think the book's notation is a bit cumbersome. I would have just used matrices and simpler letters :P $\endgroup$ Oct 6, 2018 at 17:06
  • $\begingroup$ Don't worry, it's everything ok :) $\endgroup$
    – EvaMGG
    Oct 6, 2018 at 17:12

I just had a quick chat with Prof. Montiel and it seems to be a typo in the text. Essentially, you should substitute $det (\Phi + \Psi)$ by $det (\Phi - \Psi)$.

This was the change that one could spot in the proof by Chern (page 87) that I pointed out in the comments. The difference between the proofs amounts to the fact that Montiel-Ros was taken from an article he didn't remember (he mentionned two mathematicians: the Spanish Santaló and a German called something like "Fechlossen").

I think that reusing the valuable answer by @TedShifrin you can now reconstruct what the lemma should look like and more importantly, you will be able to use it in the proof of Cohn-Vossen's theorem.

Let me know if this works out for the proof.


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