0
$\begingroup$

Let $M$ be a compact differentiable manifold and $\Delta : C^{\infty}(M) \to C^{\infty}(M)$. I am reading a lemma that states in some point of the proof:

On a compact manifold, the Laplacian has finite kernel (it is easy from Hodge theory). Even more, all non-zero eigenvalues of $\Delta$ are positive and form a simple specter, so we can pick the smallest one.

What does mean this sentence? What does mean a simple specter? Why can we pick the smallest eigenvalue?

  • 1
    I wouldn't use "In particular" because what follows is not related to the fact the Laplacian has a finite dimensional kernel. It seems that the author is trying to say that the Laplacian has a **discrete** spectrum (so the eigenvalues are isolated). Note that it is **not true** that the Laplacian has a simple spectrum (the eigenvalues can definitely be degenerate). – 2017-02-03
  • 1
    @levap, thank you so much! You are totally right, in fact what the laplacian has is a discrete spectrum! Thanks – 2017-02-03

1 Answers 1

1

The spectrum of a linear operator is said to be simple when there are no degenerate eigenspaces; that is, more than one eigenvector associated to a given eigenvalue.