How to calculate a Fredholm index numerically - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T09:54:40Z http://mathoverflow.net/feeds/question/73603 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/73603/how-to-calculate-a-fredholm-index-numerically How to calculate a Fredholm index numerically jjcale 2011-08-24T19:37:33Z 2011-08-25T07:02:04Z <p>How can one calculate the index of a Fredholm operator numerically ?</p> <p>In numerically calculations one uses always finte dimensional spaces. But linear operators on finite dimensional spaces have always index zero. </p> http://mathoverflow.net/questions/73603/how-to-calculate-a-fredholm-index-numerically/73635#73635 Answer by Helge for How to calculate a Fredholm index numerically Helge 2011-08-25T07:02:04Z 2011-08-25T07:02:04Z <p>The two key properties of the Fredholm index are</p> <ul> <li>It is a (norm)-continuous function from the bounded linear operators to the integers. In particular, if $A$ is a Fredholm operator, then there exists $\delta > 0$ such that for $\|A - B\| &lt; \delta$, we have $index(A) = index(B)$. This tells you that you can approximate your problem. </li> <li>The Fredholm index doesn't see compact perturbations. So if $A$ is Fredholm and $K$ is compact, then $index(A +K ) = index(A)$. This tells you that you cannot do naive computations like picking some finite orthonormal set $\psi_{j}$ with $j=1,\dots,N$ and hope that the $N \times N$ matrix $$A_{j,k} = \langle \psi_j, A \psi_k\rangle$$ tells you anything about the Fredholm index of $A$.</li> </ul> <p>So you will now need to do something smarter. This is possible in many particular cases, for example for Toeplitz operators. The first property allows one to reduce the computation of the index to the computation of the winding number of a polynomial. Or the Atiyah--Singer index theorems reduces computing the index to some topological information ...</p> <p>So to get a more meaningful answer, you will need to be more specific about the problem.</p>