Skip to main content
added 105 characters in body
Source Link
Bernie
  • 1k
  • 5
  • 8

I cannot comment due to reputation and haven't done the computations, but I see two possible problems:

  • usually one has a spectral sequence $(R^if_{*})(\mathcal{E}xt^j(M,N)) => \mathcal{E}xt_f^{i+j}(M,N)$ because $f_{*}\mathcal{H}om(M,-)$ is a composition of functors.

  • $Ext^2(F,F)=H^2(\mathcal{H}om(F,F))$ only holds if $F$ is locally free, due to local to global spectral sequence for $Ext$.

What may be helpful: there is a base change theorem for relative Ext-sheaves, ma be this can help you

I cannot comment due to reputation and haven't done the computations, but I see two possible problems:

  • usually one has a spectral sequence $(R^if_{*})(\mathcal{E}xt^j(M,N)) => \mathcal{E}xt_f^{i+j}(M,N)$ because $f_{*}\mathcal{H}om(M,-)$ is a composition of functors.

  • $Ext^2(F,F)=H^2(\mathcal{H}om(F,F))$ only holds if $F$ is locally free, due to local to global spectral sequence for $Ext$.

I cannot comment due to reputation and haven't done the computations, but I see two possible problems:

  • usually one has a spectral sequence $(R^if_{*})(\mathcal{E}xt^j(M,N)) => \mathcal{E}xt_f^{i+j}(M,N)$ because $f_{*}\mathcal{H}om(M,-)$ is a composition of functors.

  • $Ext^2(F,F)=H^2(\mathcal{H}om(F,F))$ only holds if $F$ is locally free, due to local to global spectral sequence for $Ext$.

What may be helpful: there is a base change theorem for relative Ext-sheaves, ma be this can help you

Source Link
Bernie
  • 1k
  • 5
  • 8

I cannot comment due to reputation and haven't done the computations, but I see two possible problems:

  • usually one has a spectral sequence $(R^if_{*})(\mathcal{E}xt^j(M,N)) => \mathcal{E}xt_f^{i+j}(M,N)$ because $f_{*}\mathcal{H}om(M,-)$ is a composition of functors.

  • $Ext^2(F,F)=H^2(\mathcal{H}om(F,F))$ only holds if $F$ is locally free, due to local to global spectral sequence for $Ext$.