MInvCG has hard-coded isign internally, is this intentional?

[kostrzewa]

The multi-shift CG has isign hard-coded to +1 and -1 internally, is this intentional and sensible?

[martin-ueding]

Since the operator() takes an isign argument, I'd say that this is not sensible. The test case only tests one isign. The hard-coded default might be implicitly used in Chroma, so one has to be careful when actually using the isign.

[kostrzewa]

@bjoo Can we safely switch this to +-isign throughout without affecting you?

[bjoo]

Hi All, I am just catching up with this. Let me see if I understand it correctly: i) operator has an isign argument ii) MinvCG ignores this and hardcodes +1,-1 iii) You’d like to change it to use +isign, -isign If so the only thing this will do is solve with M^\dag M — current (?) if isign is 1 M M^\dag if isign is -1 The point of the current hard coding was to ensure we wolve with an explicitly hermitian positive operator: M^\dagger M since this is shifted CG, and I believe this is primarily what is needed for force terms. I have not planned on using this e.g. to solve for pops. The +/- isign switch should switch to using the Normal Operator. M M^\dagger, is it desirable for you to solve this system? i.e. ( M M^\dagger + shift_i ) psi_i = chi ? Best, B

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On Jul 7, 2017, at 11:21 AM, Bartosz Kostrzewa ***@***.***> wrote: @bjoo <https://urldefense.proofpoint.com/v2/url?u=https-3A__github.com_bjoo&d=DwMCaQ&c=lz9TcOasaINaaC3U7FbMev2lsutwpI4--09aP8Lu18s&r=SC-qvz5njMoFH6cliT5XZQ&m=4bApWvhTWwyPJjWTHiiYT6-py6QvsYg0oEsJ4XLosG0&s=wCMFpk-r6qDj1lBDabvHLDjgNFnKzjqp_vUD78GXBLI&e=> Can we safely switch this to +-isign throughout without affecting you? — You are receiving this because you were mentioned. Reply to this email directly, view it on GitHub <https://urldefense.proofpoint.com/v2/url?u=https-3A__github.com_JeffersonLab_qphix_issues_89-23issuecomment-2D313632435&d=DwMCaQ&c=lz9TcOasaINaaC3U7FbMev2lsutwpI4--09aP8Lu18s&r=SC-qvz5njMoFH6cliT5XZQ&m=4bApWvhTWwyPJjWTHiiYT6-py6QvsYg0oEsJ4XLosG0&s=FNs1sZhJpyl4MIK5y0SuzAuE7X0HiYkCzEVx2YdLiPU&e=>, or mute the thread <https://urldefense.proofpoint.com/v2/url?u=https-3A__github.com_notifications_unsubscribe-2Dauth_AAj16LVMnSs69b6TfV0fEbeQW1B-2De8UUks5sLfi3gaJpZM4OQr4x&d=DwMCaQ&c=lz9TcOasaINaaC3U7FbMev2lsutwpI4--09aP8Lu18s&r=SC-qvz5njMoFH6cliT5XZQ&m=4bApWvhTWwyPJjWTHiiYT6-py6QvsYg0oEsJ4XLosG0&s=rUZTMLcOXtvQGojuxW1AoM2XfoP1gR0CkV8zyd3AoGA&e=>.

[kostrzewa]

Hi Balint,

sorry for taking so long to reply to this, it has been an unusually busy week.

The answer to your question is that yes, we solve

( M M^\dagger + shift_i ) psi_i = chi

The reason for this is basically one of convention. In tmLQCD, everything uses the hermitian Wilson Dirac operator plus a twisted quark mass (without gamma_5 then), so we have something like

Q_{+} = \gamma_5 ( M_w + i \mu \gamma_5 ) = \gamma_5 M_w + i\mu
Q_{-} = Q_{+}^\dagger = \gamma_5 ( M_w - i \mu \gamma_5 ) = \gamma_5 M_w - i\mu

And for reasons of convention, our nf=2 action is

exp{- \phi^dagger ( Q_{+} Q_{-} )^{-1} \phi }

where translating between QPhiX and tmLQCD, we have to insert appropriate factors of \gamma_5 but everything goes through smoothly with the definitions that we've made with the twisted mass addtitions to QPhiX as long as we solve the "isign=-1" system. ( M = M_w + i \mu \gamma_5, M^dag = M_w^dag - i \mu\gamma_5)

Now the situation for nf=2 (degenerate) also translates to the non-degenerate doublet or a single flavour of Wilson quarks and so we need to solve the shifted M.M^dagger system rather than the shifted M^dagger.M system.