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.

Let $I \in C^1(\mathbb{R}^n,\mathbb{R})$ be an even functional. There is a claim that then the restriction to the sphere has the following form for the Frechet derivative $$I|_{S^{n-1}}'(u) = I'(u)-(I'(u),u)u$$ I don't see why this is true. Could you explain please?

share|improve this question

1 Answer 1

The Fréchet derivative does not make sense here, because $I|_{\mathbb{S}^{n-1}}$ is not a function defined on an open set of a linear space. I think what is meant here is the gradient of $I|_{\mathbb{S}^{n-1}}$ as a function on the Riemannian manifold $\mathbb{S}^{n-1}$. That is a vector field, i. e. a section of the tangent bundle of the sphere, so embedded in the tangent bundle of $\mathbb{R}^n$, at each point $u \in \mathbb{S}^{n-1}$ it has to be orthogonal to $u$. In fact, it is the orthogonal projection of the gradient of $I$ in $\mathbb{R}^n$ along $u$, and that is what the formula is supposed to mean.

share|improve this answer
    
Do you mean, the gradient of I (that is, not the gradient of $I|_{S^{n-1}}$) as a function on $S^{n-1}$, since you just said that the derivative does not make sense here? –  Euler....IS_ALIVE Jan 25 '13 at 18:15
    
also if this expression is orthogonal to $u$, then that means that we should (evaluating at $u$) that $(I'(u),u) - (I'(u),u)u = 0$. This doesn't seem to be the case. –  Euler....IS_ALIVE Jan 25 '13 at 22:07
    
@Euler....IS_ALIVE Frechet derivative (a concept of functional analysis) does not make sense for $I_{|S^{n-1}}$, but the gradient (concept of Riemannian geometry) does. –  user53153 Jan 25 '13 at 22:51
    
@Thomas I would be surprised, since the author makes no reference to Riemannian geometry. This is in fact from a PDE book. The goal is to prove that under the conditions for $I$ above, then $I|_{S^{n-1}}$ has at least $n$ distinct pairs of critical points, apparently first proved by Ljusternik and Schnirelmann. This link is not the book I am studying, but it is by the same author with the same claim tinyurl.com/aq27j92 p.153-154 –  Euler....IS_ALIVE Jan 25 '13 at 23:02
1  
@Euler....IS_ALIVE It is orthogonal to $u$: $(I'(u)-(I'(u),u)u,u) = 0$, since $(u,u) = 1$. I guess the author implicitely parametrizes the sphere as a submanifold of $\mathbb{R}^n$, concatenates the parametrization with $I$. Then he can take the gradient, but has to push it forward to $\mathbb{R}^3$ via the derivative of the parametrization. –  Thomas Jan 27 '13 at 14:50

Your Answer

 
discard

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.