Home ยป Coloring the Odd Distance Graph

Flaw

On May 8th, 2008 JSteinhardt says:

Actually, it appears there is a flaw with the below proof. The spectral bound on the chromatic number assumes a measurable coloring, as we want to say the following:

$$2(\chi-1)\lambda_{\min}||f||^2 & = & \sum_{i,j=1}^{\chi} \lambda_{\min}||f_i-f_j||^2$$ $$\leq \sum_{i,j=1}^{\chi} \langle f_i-f_j, B(f_i-f_j) \rangle$$ $$= \sum_{i,j=1}^{\chi} \langle f_i,Bf_i \rangle + \langle f_j,Bf_j \rangle - 2\langle f_i,Bf_j \rangle$$ $$= -2\sum_{i,j=1}^{\chi} \langle f_i,Bf_j \rangle$$ $$= -2\langle\sum_{i=1}^{\chi} f_i,B(\sum_{i=1}^{\chi} f_i\rangle$$ $$ = -2\langle f,Bf \rangle$$ $$ = -2\lambda ||f||^2$$

but this assumes that each $ f_i $ is Lesbegue integrable, which in turn requires measurable coloring classes.

Reply

Comments are limited to a maximum of 1000 characters.
More information about formatting options