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$M$ compact manifold. Let $\lambda$ be an eigenvalue for the Dirac operator of multiplicity greater than 2. I'm interested in showing the existence of two linearly independant eigenspinors $u$ and $v$ of eigenvalue $\lambda$, with $\|u\|_2=\|v\|_2=1$ and so that there exists a point $x_0 \in M$ for which $\|u(x_0)\|_{x_0} \neq \|v(x_0)\|_{x_0}$. Trying to show this by contradiction, I assumed that $\|u(x)\|_x = \|v(x)\|_x$ for all $x \in M$. I can prove the desired result provided there be two linearly independant eigenspinors of eigenvalue $\lambda$ with $\langle u,v \rangle_2 \neq \frac{\langle u(x),v(x) \rangle_x}{|u(x)|_x^2}$ for some $x \in M$.

My question is:

Is it possible that any two normalized linearly independant eigenspinors $u$,$v$ in such $\lambda$ eigenspace satisfy both $$\|u(x)\|_x = \|v(x)\|_x \quad \quad \text{and} \quad \quad\langle u,v \rangle_2 =\frac{\langle u(x),v(x) \rangle_x}{|u(x)|_x^2} \quad \text{for all } x \in M \;?$$

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1 Answer

The Dirac operator on $M$ is self-adjoint (or skew-adjoint, depending on your conventions). So the $\lambda$-eigenspace must be spanned by eigenvectors. Does this not solve your problem straight away?

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I don't see how. I'm already assuming that they are eigenvectors. –  user16436 Jan 27 '12 at 4:04
    
What do you understand by "multiplicity"? Normally one would mean the dimension of the (generalized) eigenspace. Then you're done. –  Bob Yuncken Jan 27 '12 at 12:19
    
Let's see. If $u$ and $v$ are two orthonormal linearly independant eigenvectors of Dirac, then $\|u\|_2=\|v\|_2$ and $\langle u,v \rangle _2=0$. If in addition I ask that for every point in the manifold the norms in the fiber are equal ($\|u(x)\|_x=\|v(x)\|_x)$), it won't be always true that $$\langle u(x), v(x) \rangle_x =0 \quad \forall x\in M.$$ Indeed, I expect it to be very unlikely. –  user16436 Jan 27 '12 at 15:37
    
Thank you very much for giving some thought to my problem. –  user16436 Jan 27 '12 at 15:38
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