About the original variational LDA algorithm

Ok, let me take a stab at this:

why is inference used every time the document e step is computed (ie, why does lda-estimate.c:doc_e_step call lda-inference.c:lda_inference?)

So instead of going through steps 3 & 4 above sweeping across all documents, you can instead do that on a per-document basis since the documents are independent conditional on the topics.

why in lda-inference.c:lda_inference is phi_nk updated as digamma(gamma_k) + log(Pk(w)) instead of Pk(w) exp(digamma(gamma_k) - digamma(sum(gamma))), as equation 16 from the paper would lead one to conclude? I see that it is computing log(phi_nk) instead of phi_nk, but there seems to be a missing digamma factor. Can he omit that because it disappears after sum-normalizing the phi variables?

Yes, the last term is constant so when phi is normalized, it is irrelevant.

lda-inference.c:lda_inference seems to do many passes until convergence. Why should it converge when you're just estimating the model? Don't these extra iterations screw up the estimations of phi and gamma?

Again, the order you do steps 3 and 4 is just a heuristic. It's debatable which approach is actually better, but as above, because of independence, this is totally valid.

phi seems to be reestimated using a default gamma (in lda-inference.c:lda_inference) equal to alpha*nwords/ntopics, and gamma seems to be reestimated after this (for which reason?) as this_value + count[w] * (phi_nk - oldphi_nk) (after normalizing phi), and this estimate is reupdated for each word in each document, during inference. Doesn't this mean that the phi for different words will see different values of the gamma? And why not following equation 17 from the paper?

This is coordinate ascent, so again the order is arbitrary. You can update gamma, then update all the phis, then update gamma, then update all the phis, etc. Or you can update gamma, update phi_1, update gamma, update phi_2, and just repeat. I think this approach here gives you empirically better convergence, but it's a toss up really.