ag.algebraic geometry - When two k-varieties with the same underlying topological spaces isomorphic?

[This is a situation where things have been rewritten several times in response to an ongoing discussion. Let me try to reconstruct some of the temporal sequence here.]

ROUND ONE:

Ben's answer gives one way that your statement can fail: $Y$ can be singular and $f$ can be a birational morphism which does not induce an isomorphism of local rings at at least one of the singular points.

Here is something else that can go wrong: in positive characteristic, $f$ can be purely inseparable, e.g. the $p$-power Frobenius map.

ROUND TWO

I edited my response to point out that Ben's example has a nonreduced fiber over the singular point. I also said "I think" that mine does not, but this was pointed out by Kevin Buzzard to be false. [Or rather, the statement is false. I truly did think it was true for a little while.]

I also suggested that the following modification might be true:

Suppose $X$ and $Y$ are geometrically irreducible and $Y$ is nonsingular (together with all of the questioner's hypotheses, especially reducedness of the fibers!). Then if $f:X rightarrow Y$ is a bijective morphism with reduced fibers, it is an isomorphism.

ROUND THREE

I typed up a counterexample over an imperfect ground field when the varieties are not geometrically integral (g.i. = the base change to the algebraic closure is reduced and irreducible: the reduced business has to be taken more seriously when the ground field is imperfect, since taking an inseparable field extension can introduce nilpotent elements). But Kevin Buzzard posted a simpler counterexample, so I deleted my answer.

ROUND FOUR

Kevin's answer also includes a beautifully simple example to show that the question is false even over $mathbb{C}$ without some nonsingularity hypotheses: use nodes instead of cusps and remove one of the preimages of the nodal point.

I still wonder if my attempted reparation of the statement above is correct.

ag.algebraic geometry - Decomposition theorem and blow-ups

In this example we have $p : X to Y$ and we may assume, wlog, that $X$ is isomorphic to the total space of the normal bundle to the surface, and $p$ is the contraction of the zero section.

Then, by the Deligne construction, $IC(Y) = tau_{le -1} j_* mathbb{Q}[3]$, where $j : Y^0 hookrightarrow Y$ is the inclusion of the smooth locus (which is isomorphic to $X^0$ the complement of the zero section in $X$).

In order to work this out, we can use the Leray-Hirsch spectral sequence

$E_2^{p,q} = H^p(S) otimes H^q(mathbb{C}^*) Rightarrow H^{p+q}(X^0)$

this converges at $E_3$ and we get that the degree 0, 1 and 2 parts of the cohomology of $X^0$ is given by the primitive classes in $H^i(S)$ for $i = 0, 1, 2$. Note that this is everything in degrees 0 and 1, but in degree two the primitive classes form a codimension one subspace $P_2 subset H^2(S)$.

The Deligne construction above, gives us that $IC(Y)_0 = H^0(S)[3]
oplus H^1(S)[2] oplus P_2[1]$.

(This is a general fact: whenever you take a cone over a smooth projective variety, the stalk of the intersection cohomology complex at 0 is given by the primitive classes with respect to the ample bundle used to embed the variety. This follows by exactly the same arguments given above.)

Then the decomposition theorem gives

$p_* mathbb{Q} = mathbb{Q}_0[1] oplus ( IC(Y) oplus H^3(S) ) oplus H^4(S)[-1]$.

EDIT: fixed typos pointed out by Chris.

gt.geometric topology - Proof of the Reidemeister theorem

I taught knot theory last semester and ran into the same problem. I looked in every book I could get my hands on, and could not find an undergraduate level proof. In the end, I wrote up my own notes (which I would be happy to scan when I get back into the office). The key ideas for the case-by-case analysis are in the book "Knots, links, braids, and 3-manifolds" by Prasolov and Sosinsky. I also found Louis Kauffman's book "On knots" to be helpful. There are two lemmas I could not find anywhere: (1) the general position argument, which says that there is a nice projection and (2) the argument which says that you can find a general projection so that the associated diagram is equivalent to the original diagram (most books skip this issue). The point of the second lemma is that it is not enough to show that there exist two projections that differ by Reidemeister moves, rather, you want to show that the two given diagrams differ by Reidemeister moves.

nt.number theory - Bounds on $sum {n ((alpha n))}$ where $((x))$ is the sawtooth function

Here is an approach which may give some better estimates for particular values of $alpha$:

$$sum_{i= 1}^N i((ialpha)) = sum_{i=1}^N sum_{j=i}^N ((j alpha)) = sum_{i=1}^N sum_{k=0}^{N-i}((Nalpha -kalpha))$$.

So, if you can estimate

$$sup_{substack{xin (0,1) \ M le N}} bigg|sum_{k=0}^{M} ((x-kalpha))bigg|,$$

then you can crudely multiply by $N$ to get an estimate for $|sum i((ialpha))|$.

Specifically, my guess is that for quadratic irrationals $alpha$, there is an upper bound for

$$bigg|sum_{k=0}^M ((x-kalpha))bigg|$$

which is $O (log M)$, which would give you a bound of $O(N log N)$, and more generally that there is a bound in terms of the coefficients of the simple continued fraction for $alpha$, so that if those are bounded, then you still get $O(N log N)$.

For the particular value $phi = (sqrt5 + 1)/2$, $sum_{i=0}^{M} ((i phi))$ has logarithmic growth $c + (5sqrt5 - 11)/4 log_phi M$ (achieved at indices in the sequences A064831 (+) and A059840 (-)), which suggests that $sup sum_{i=0}^M ((x-iphi))$ also has logarithmic growth, which would give an $N log N$ bound for the sum.

---

In the opposite direction, for all $alpha notin frac 12mathbb Z$, $$limsup bigg(log_N bigg|sum_{i=0}^N i((ialpha))bigg|bigg) ge 1$$ since there are terms proportional to $N$.

The sum can be greater than $N^{2-epsilon}$ infinitely often by choosing $alpha$ so that it is extremely well approximated by infinitely many rational numbers. When $alpha$ is very closely approximated by $p/q$, then for $N$ a small multiple of $q$ (where "small" is relative to how well $p/q$ approximates $alpha$), about $1/q$ of the terms can be moved past integers with a small perturbation of $alpha$ to $alpha'$, which causes a jump of about $N^2/q$ in the sum. So, either the sum for $alpha$ or $alpha'$ is large. We can choose a sequence $p_n/q_n$ which converges to an $alpha$ which produces large sums infinitely often, so that for these $alpha,$
$$limsup bigg(log_N bigg|sum_{i=0}^N i((ialpha))bigg|bigg) = 2$$.

Homological Algebra for Commutative Monoids?

I used to think about this problem in relation to a chain theory for bordism ( as mentioned by Josh Shadlen above).
The problems you have with monoids is first and foremost that the category is not balanced. That means that you can have an epimorphism that is also a monomorphism but NOT an isomorphism. eg. the inclusion of N --> Z.
Subsequently most constructions that you would like to make - notably short exact sequences and the snake lemma - fail at some level.
I made a few notes on this as part of my investigation into bordism theory / homework assignment
here.

In this case, we have complexes of free abelian monoids whose homology takes it's values in abelian groups and yet, the long exact sequence does not come from a short exact sequence of monoid complexes.

My references include:

[Bau89] Friedrich W. Bauer. Generalised homology theories and chain complexes. Annali di Mathematica pura ed applicata, CLV:143–191, 1989.

[Bau95] Friedrich W. Bauer. Bordism theories and chain complexes. Journal of Pura and Applied Algebra, 102:251–272, 1995.

[BCF63] R.O. Burdick, P.E. Conner, and E.E. Floyd. Chain theories and their derived homology. Proceedings of the AMS, 19(5):1115–1118, Oct. 1963.

[Koc78] S. O. Kochman. A chain functor for bordism. Transactions of the American Mathematical Society, 239:167–196, 1978.

at.algebraic topology - Burnside ring and zeroth G-equivariant stem for finite G

This is from Segal's paper

Equivariant stable homotopy theory. Actes du Congrès International des Mathématiciens (Nice, 1970), Tome 2, pp. 59--63. Gauthier-Villars, Paris, 1971.

MR0423340 (54 #11319)

where he states that the equivariant stable cohomotopy of $S$ is $pi_G^{-ast}(S)=bigoplus_Kpi_{ast}^s(B(N_G(K)/K)^+)$. In particular, $pi_G^0(S)$ is the Burnside ring.

universe - What would we find if we could travel 786 trillion light years in any direction

There is one principle that says "The universe is isotropic". It means it looks the same (on large scales) from any point you look from.

This means, in turn, that if you travel 786 trillion light years instantaneously in some direction, you'll find a very differnt sky but with exactly the same general structure: stars, galaxies, clusters of galaxies, and attractors.

Altough you can NOT travel so far so fast, you can make this mind experiment:

Send a spaceship so far at (almost) speed of light with a photo camera, take a 360º photo, come back, arrive 1572 triilion years afterwards.
While awaiting for the spaceship return, wait 786 trillion light years, take a 360º photo, wait the othe 768 trillion years.

Compare the two photos. They were taken at the same "Cosmic time", one of them 768 trillion years ago, here, and the other one maybe two or three years of spaceship time ago, 768 trillion light years afar.

The photos will look the the same on the rough scale.