Alex Youcis

Berkeley, CA

ayoucis.wordpress.com

Age: 24

I am a second year graduate student at the University of California, Berkeley.

My interests are generally in the field of arithmetic geometry. More specifically, I am interested in various aspects of the Langlands program. My advisor is Sug Woo Shin.

Apr
28
revised Best Sets of Lecture Notes and Articles
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Apr
25
comment norm map and local class field theory
They are abstractly isomorphic, but the restriction map is not an isomorphism. You should, as good practice, write down why this is true, but it's analagous to the fact that :$G_{\mathbb{F}_{p^r}}\to G_{\mathbb{F}_p}$ is not an isomorphism, even though the two groups are isomorphic (the map is multiplication by $r$). You'll find that your restriction map is surjective though, which is good since the norm surjects for unramified extensions.
Apr
20
comment 'Stalk' of vanishing cycles at $k$-point
$g^\ast\mathbb{Q}_\ell=\mathbb{Q}_\ell$ identification if we just think of the $G_K$-representation $H^i(X_{\overline{K}},\mathbb{Q}_\ell)$ as the $\overline{\eta}$-stalk (corresponding to the choice of $\overline{K}$) of the smooth $\mathbb{Q}_\ell$-sheaf $R^if_\ast\mathbb{Q}_\ell$ (where $f$ is the structure morphism $X\to\mathrm{Spec}(K)$). Is there something similar here? Sorry for beating a dead horse, and thanks!
Apr
20
comment 'Stalk' of vanishing cycles at $k$-point
I apologize for being annoying. As you can see, this is a purely silly notational question. So, we want a canonical isomorphism $\overline{g}(\overline{x}}=\overline{x}$. I guess this is because $\overline{g}\circ\overline{x}=\overline{x}\circ\overline{g}$ since $\overline{x}$ lies over a $k$-point (where on the right $\overline{g}$ acts on $\mathrm{Spec}(\overline{k})$). And then we just identify this RHS with $\overline{x}$ via $\overline{g}^{-1}$? Is that right? Also, is there a way of phrasing this without having to make such an identification? For example, we can ignore the
Apr
20
revised 'Stalk' of vanishing cycles at $k$-point
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Apr
20
comment 'Stalk' of vanishing cycles at $k$-point
Just to be explicit. Literally taking stalks gives me an isomorphism $\sigma(g)_{\overline{x}}:(\overline{g}^\ast (R\Psi\overline{\mathbb{Q}_\ell}))_{\overline{x}}\to (R\Psi\overline{\mathbb{Q}_\ell})_{\overline{x}}$. But, this is just an isomorphism $(R\Psi\overline{\mathbb{Q}_\ell})_{\overline{g}(\overline{x})}\to (R\Psi\overline{\mathbb{Q}_\ell})_{\overline{x}}$. NOT an isomorphism $(R\Psi\overline{\mathbb{Q}_\ell})_{\overline{x}}\to (R\Psi\overline{\mathbb{Q}_\ell})_{\overline{x}}$ which is what I would want to have an action. Like I said, maybe this is just some 'canonical identification
Apr
20
comment 'Stalk' of vanishing cycles at $k$-point
of $\mathrm{Gal}(\overline{\eta}/\eta)$ preserves the geometric point. Of course, it doesn't though. Given $g\in\mathrm{Gal}(\overline{\eta}/\eta)$ we actually have a map between the $\overline{g}(\overline{x})$ and the $\overline{x}$ stalk of $R\Psi\overline{\mathbb{Q}_\ell}$. I think the confusion might be in some canonical identification between these stalks, but I'm not sure (e.g. how when we define the $G_K$ action on $H^i(X_{\overline{K}},\mathbb{Q}_\ell)$ we make the 'canonical' identification between $\overline{Q}_\ell$ and $g^\ast\overline{\mathbb{Q}_\ell}$).
Apr
20
comment 'Stalk' of vanishing cycles at $k$-point
I think your first paragraph is what I said. The choice of $\overline{\eta}$ gives me the choice of $\overline{s}$ (by normalizing $S$ in $\overline{\eta}$ and taking the residue field) which is why there is a canonical $\bar{x}\in X_{\bar{s}}(\overline{k})$ associated to $x$. But, what you didn't answer (unless I am misunderstanding you) is why this actually gives me a well-defined action of $\mathrm{Gal}(\overline{\eta}/\eta)$ on $(R\Psi\overline{\mathbb{Q}}_\ell)_{\overline{x}}$. My worry is that we should get the action just because the compatible action
Apr
20
awarded Editor
Apr
20
revised 'Stalk' of vanishing cycles at $k$-point
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Apr
20
asked 'Stalk' of vanishing cycles at $k$-point
Apr
12
awarded Necromancer
Apr
10
comment Rank 1 Azumaya algebra
What does constant rank $1$ mean? That etale locally on $X$, $A$ is $\mathrm{Mat}_1(\mathcal{O}_U)=\mathcal{O}_U$? The map $\mathrm{Br}(X)\to \mathrm{Br}'(X)$ is injective (it's just the limit of the inclusions $H^1(X,\mathrm{PGL}_n)/H^1(X,\mathrm{GL}_n)$), so the image is trivial if and only if your original algebra is trivial!
Mar
15
comment A simple question about rational Hodge conjecture
I'm not quite sure, but perhaps it's because we have Lefschetz $(1,1)$?
Mar
12
comment Why are finite group schemes usually assumed flat?
important, I feel like, often times, just masks the true nature of theorems/notions. So, in case I am understanding you, if you assume that everything in sight is Noetherian, do you feel satisfactorily about the definitions?
Mar
12
comment Why are finite group schemes usually assumed flat?
I'm confused by your question. I think you're saying that you understand why you want to assume locally free of finite, constant rank, right? Because then you can define the order of your finite group scheme. So, you're asking why not just say locally free of finite constant rank? Probably because, as you observed, for finite morphisms to Noetherian things, this is the same thing (assuming connected base). So, I think the answer to both, and I mean this sincerely: don't pay attention to Noetherian hypotheses. Wondering why you need this or that to be Noetherian, while theoretically
Mar
12
comment $\mathbb{G}_a$ or $\mathbb{G}_m$ as subgroups of Affine Algebraic Groups
If you're working over $\bar{k}$, then this is equivalent to "does every affine algebraic group contain an integral smooth subgroup of dimension $1$". Somehow, this seems even harder though!
Mar
12
comment For a $k$-morphism $X \to Y$ to be determined by $X(\overline{k}) \to Y(\overline{k})$, does $X$ really have to be _geometrically_ reduced?
@KReiser No butting in. I couldn't have said it better myself :) You should ping c_c_chaos though so he knows you responded.
Mar
11
revised For a $k$-morphism $X \to Y$ to be determined by $X(\overline{k}) \to Y(\overline{k})$, does $X$ really have to be _geometrically_ reduced?
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Mar
11
comment Hartshorne's Algebraic geometry Chapter III ex. 9.10
Very nice example Georges! I guess what you wrote is fairly intuitive. You have a family of conics which, fiber by fiber are of course trivial (they have a zero), but to create a global trivialization you'd have to choose a family of rational points which, as you mentioned there is none.
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