Today was the first day after my prelims that I met Dr. Shahidi. He earlier advised me to take a break after the advanced topics so I could start research afresh. I also recently found his Wikipedia page (albeit a stub) – http://en.wikipedia.org/wiki/Freydoon_Shahidi . A more relevant Wiki-link about his work is – http://en.wikipedia.org/wiki/Langlands%E2%80%93Shahidi_method I hope one day I’ll edit these pages to add sources and more relevant material.

Our conversation was quite pleasant. He had some things in mind he wanted me to work on. He told me to go over “L-packets” and “A-packets” and other technical stuff; gave references where I could read about it. We discussed about my general PhD goal. “Your (research) problem should not be too difficult to be unable to solve”, he said. On asking if I’d like my work to be “Algebra or Analysis”, I instantly replied “Algebra!” although I should know that Number Theory uses tools from all branches of mathematics (including PDE )

I have known Shahidi for his dry sarcastic wit and today’s conversation ended with a remarkable quip. He told me, “Keep me informed, don’t run away!”

Recently (22 March 2013) I took my advanced topics exam. In spirit of the Princeton Generals, I wrote out my interview questions for use to anyone in these special topics.

Topics: linear algebraic groups, class field theory

Committee: Freydoon Shahidi (chair), David Goldberg

Linear algebraic groups:

Goldberg decided to start with Linear algebraic groups. What is meant by ‘split’? (I told I had prepared only the algebraically closed case. They were okay). What is parabolic? If Q is parabolic in G and P is parabolic in Q, prove that P is parabolic in G. (Went totally blank. They gave hints – use the other equivalent definition).

State Bruhat decomposition. (I started defining terms – maximal torus, root system, Weil group etc.) How is the Weyl group related to the torus? (Another moment when I blanked out). How does W act on T? (By conjugation. So W is the quotient of the normalizer of the maximal torus by its centralizer). When is the centralizer equal the torus? (When it is maximal). Shahidi objected and corrected me twice when I wrongly pronounced “vile” for Weyl group – correct is “veil”. There is another group after Andre Weil.

Bruhat decomposition for GL_n. What is it’s Weyl group? (S_n). What is the length of an element? (minimal length of the decomposition in terms of reflections). What is the polynomial the big cell satisfies? (I had prepared this one. Told the answer and that the proof goes by induction but Shahidi was not satisfied. He told something I didn’t quite understand. I gave the decomposition explicitly and he told it generalizes the well-known LU decomposition to classical groups).

What is the critical step in the classification of semisimple groups of rank 1? (Uses Bruhat decomposition). Give a sketch of proof. (I couldn’t. But turns out, they really wanted me to state the theorem. SL_2 and PSL_2).

Given that the normalizer of a parabolic is itself, show that it must be connected. (Missed a step in that one can choose the conjugating element inside the connected component).

[ That was all about algebraic groups. Nothing about Dynkin diagrams, finding parabolics, simple connectedness and fundamental groups of classical groups. ]

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Number Theory:

Class group of Q(\sqrt -7). (Minkowski bound works).

Class group of Q(\sqrt 21). (I saw that 25 – 21 = 4 and started considering the factorization of the prime 2 above using quadratic reciprocity but got it wrong. It took me almost 10 grueling minutes to figure out that 21 was not a prime! Embarrassment).

Define the Hilbert symbol. State the norm condition. Prove the bi-multiplicativity property. (Consequence of a group homomorphism). State and prove Hilbert’s reciprocity. (I had prepared a classical proof from Serre but he insisted on proving via Class field theory. Proved that the local p-Artin maps glue to give a global Artin map and the theorem follows elegantly).

Goldberg asked if I knew anything about the generalized symbol. (I stated and mentioned the skew-symmetry property). He said it’s not in your syllabus anyways.

Show that p = a^2 + ab + b^2 precisely when p is 1 (mod 3). (I was looking at the field of cube roots of unity but was blundering some norm calculation. After many frustrating minutes for everyone, it was discovered that the problem was wrong and should have p = a^2 – ab + b^2. I commented, “I have been doing this calculation since tenth grade” on which Prof. Shahidi quickly retorted, “still you haven’t memorized it!”

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The exam lasted for 90 minutes. They told me to wait outside and after a stressful 5-minute period for me, they came out and congratulated me.

Shahidi: Good, congratulations.

Me: Thanks.

Shahidi: Don’t see me for a month now.

Me: Yeah, the past two weeks were tough on me too.

Shahidi: Let us not meet each other for a while now. (All smile in acknowledgement).

Recently, I’ve been preparing for my advanced topics exam at Purdue and have been going over some algebraic number theory. The question in consideration was to find the Hilbert class field (i.e., the maximal abelian unramified extension) of ${\mathbb Q(\sqrt{-6})}$. The problem boiled down to determining how a prime split in an extension, which reduced to finding the ring of integers of a number field. The usual method (following Kummer) is to find an algebraic integer that generates the ring of integers of the extension over the ring of integers of the base field. But such an integer is not guaranteed to exist. I posted this on math.stackexchange and got an interesting answer in the form of a comment which I now blog.

Theorem: Let ${K}$ and ${L}$ be linearly disjoint number fields, i.e.,

$\displaystyle K \otimes_{\mathbb Q} L \rightarrow KL$

is an isomorphism. Then, ${D_{KL} | D_K^s D_L^r}$, where ${D_K}$ denotes the discriminant of ${K}$ over ${\mathbb Q}$, etc. and ${r = [K:\mathbb Q]}$ and ${s = [L : \mathbb Q]}$. Moreover, if ${D_K}$ and ${D_L}$ are relatively prime over ${\mathbb Z}$, then ${D_{KL} = D_K^s D_L^r}$ and ${\mathcal O_K \otimes_{\mathbb Z} \mathcal O_L \hookrightarrow \mathcal O_{KL}}$ is an isomorphism. So, the basis of ${\mathcal O_{KL}}$ can be obtained from the bases of ${\mathcal O_K}$ and ${\mathcal O_L}$ in a natural way.

Proof:
Let ${M}$ be the image of

$\displaystyle \mathcal O_K \otimes_{\mathbb Z} \mathcal O_L \hookrightarrow \mathcal O_{KL}.$

Then ${M}$ is an order of ${\mathcal O_{KL}}$ (i.e., a subring of the ring of integers). If ${x = ab \in M}$ is the image of a pure tensor ${a \otimes b \in \mathcal O_K \otimes \mathcal O_L}$, then the trace form on ${KL}$ acts on ${x}$ by ${T_{KL}(ab) = T_K(a).T_L(b)}$. (For a proof of this, go to the Galois closure and use the canonical isomorphism ${\text{Gal}(KL/\mathbb Q) \cong \text{Gal}(K/\mathbb Q) \times \text{Gal}(L/\mathbb Q)}$ given by linear disjointness). Thus, ${T_{KL} : M \rightarrow M}$ acts by ${T_K \otimes T_L}$. The matrix of ${T_{KL}}$ is then the Kronecker product of the matrices of ${T_K}$ and ${T_L}$. It follows that ${D_M = D_K^s D_L^r}$. By transitivity of discriminants (see below), ${D_{KL} | D_K^s D_L^r}$. By considering the towers ${\mathbb Q \subseteq K \subseteq KL}$ and ${\mathbb Q \subseteq L \subseteq KL}$, we also know that ${D_K^s D_L^r}$ is divisible by ${D_K^s}$ and ${D_L^r}$, hence the assertion that ${D_{KL} = D_K^s D_L^r}$. It follows in this case (again by transitivity of discriminant) that ${M = \mathcal O_{KL}}$.

Transitivity of discriminants:
Let ${K\subseteq K' \subseteq K''}$ be a finite extension of number fields and ${A, A', A''}$ be their rings of integers. Then the discriminants satisfy

$\displaystyle D_{A''/A} = D_{A'/A}^{[K'':K']}.\mathrm{N}_{ K'/ K}(D_{ A''/ A'}).$

Today’s post is about group theory. Well, to state the result, one needs to know just group theory but proving it requires some knowledge of root systems. And one really understands its application when studying algebraic groups. In my opinion, one can rarely say while learning mathematics that s/he will not need a particular result, or something is not useful in their research. So it’s always good to know something and connect to it later when learning something related.

The Bruhat decomposition is a decomposition of a group having a “Tits system” into double cosets. It is used to decompose $GL(n,F)$, the group of $n \times n$ invertible matrices over a field $F$, or any reductive connected closed subgroup of this matrix group.

For a group $G$, a Tits system (named after Jacques Tits) is a triple $(B, N, I), \, B, N$ subgroups of $G$ called a $BN-$pair and $I$ is a set such that the following axioms hold:

1. $T := B \cap N \unlhd N$,
2. $I$ is a set of generators of $W := N / T$ (called the Weyl group) and $s^2 = 1$ for all $s \in I$,
3. If $w \in W$ and $s \in I$, then $wBs \subseteq BwsB \cup BwB$,
4. If $s \in I$, $sBs^{-1} \neq B$,
5. $B, N$ generate $G$.

Remarks:

• (1) says that $W$ is a group.
• In (3), $W$ is not necessarily a subgroup of $G$. Yet, define $wB$ as $wB := nB$, where $w=nT$ and observe the well-definedness.
• We’ll be concerned with “Bruhat cells”, C(w) := BwB. With this notation, (3) is equivalent to $C(w) C(s) \subseteq C(ws) \cup C(w)$.
• From (4), $1 \not\in I$.
• From (2), each $s \in I$ has order 2.

Theorem (Bruhat decomposition).

Let $(B, N, I)$ be a Tits system and $W = N/T$ be the Weyl group. Then,

$G = \coprod_{w\in W} C(w)$.

A (rough) sketch of the proof:

Let $G' = \cap_{w\in W} C(w)$. Prove that $G’$ is a group, and in fact a subgroup of $G$.

$G'$ contains $N$ and $B$ which generate $G$ so $G' = G$.

For disjointness, it suffices to show $w_1 = w_2$ whenever $C(w_1) = C(w_2)$.

Observe that two double cosets are either disjoint or equal.

The canonical example:

$F$ is any field and $GL(n,F)$ is the set of invertible matrices over $F$. The Borel subgroup $B$ consists of upper triangular matrices and take $N$ to be the subgroup of monomial matrices, i.e., each row and each column has a unique nonzero entry. Then $T = B \cap N \cong (F^*)^n$, which is a torus. One does however need to show that this forms a Tits system.

Recently while browsing in the Purdue Math library, I found notes of Armand Borel‘s talk titled `Algebraic groups and Arithmetic groups’. It gives an overview of algebraic groups and its applications. They are very sleek (but mathematically dense) so I decided to type them out. These notes discuss Tits systems, and their deep connections to Geometry, group theory and number theory. You can read them here.

Life is like riding a bicycle. To keep your balance, you must keep moving. — Albert Einstein

With this in my mind, I pushed myself against the winds as I completed 125 miles from Bloomington to West Lafayette in two days. It was a fundraising bike-ride called the “Bucket 100” organized by the Habitat for Humanity, an organization that builds houses for the homeless.

Let me firstly heartily thank my dear friends who helped me complete my fundraising goal. Kudos!

We set out on Saturday morning on a bus to Bloomington, IN. The night halt was 60 miles away at a school at Danville and today morning (28 Oct, Sunday) we set out for Purdue.

Tanner and me

The maximum I have ever rode on a bike was around 30 miles, around Purdue, so 60 in a day, and then 65 the next day seemed formidable. On the first day, the winds were not very strong but Bloomington is full of small hills. (In fact, the beautiful IU campus reminds me of the IITB campus with its hilly terrain). I found a biker friend in Tanner, whom I befriended in one of the practice rides. We encouraged each other to pull ourselves; at one point, we missed a turn and ended up cycling 6 miles extra. The last 10 miles were the toughest with daunting hills but the drive to reach kept us going. After arduous efforts I made it to Danville, wondering how I would make tomorrow’s 65 miles. But it was okay as I did those (dull) ‘stretching’ exercises.

The next morning (today) was much more challenging. The route from Danville to West Lafayette was a typical mid-west one. You can see nothing but farm-land for miles on end. The north (Chicago) winds have nothing – no forests, trees, houses – to stop them and we had to ride head-on into the wind. Tanner fell behind but I found myself riding with Karen, an awesome biker from near Chicago. We drafted each other from the wind alternately without which I could never have completed my ride. On the way, we passed lakes, rivers, railway-tracks, stinky pig-farms and many strange animals. At one point, a pet dog started following us even beyond a mile! We carried it to a nearby golf-club and whose owner was kind enough to keep it for a while.

Karen and I after completing the ride

The SAG vans (support and gear) stopped at regular intervals and we could replenish ourselves with energy bars, bagels, energy drinks and also chat with other bicyclists. Elaina and Marcus did a fantastic job of coordinating the entire ride – from marking the route to arranging the victory cake. It was great to get to know some cool people from Purdue and beyond!

I tracked my whole ride through my iPhone and it’s fun taking a look at the statistics. Here is the link -

Some ride photos

PS: Marcus has clicked an awesome photo of me, just as I rode toward the finish line. I will upload it as soon as he emails it to me. (Here it is…)

At the finish line

PS: The previous post on the blog was about mandibular fractures, my experience about the broken jaw from the bicycle accident. I would like to tell the distressed reader that recovering from a fracture takes time but life comes back to normal sooner than you expect!

I recently suffered from a mandible fracture and had a wired jaw for six weeks wherein I could only drink liquids. After removing the rubber-bands and braces, I went to India for the subsequent dental treatment. In this post, I describe the do’s and don’ts while recovering a mandible fracture. It is mainly a description of my experiences.

Nerve damage causes numbness – I realized that I lose sensation in some parts of the cheek. It probably is because of nerve damage because of the accident impact. Most likely it will heal within an year (mine took just a month) but there are instances of numbness never going away.

Talking with a closed mouth – In the first week, I was not able to talk clearly and it would pain after talking continuously for few minutes. By the third week, I became an expert in talking with a closed jaw. By the fifth week people couldn’t notice any abnormality unless they peeked and saw the metal braces.

A compete liquid diet – This is described in details in this post. Keep in mind that you would lose upto 10% of your body weight due to no solid intake. (In retrospect, I think it was good to lose the few extra pounds I had always wanted to. But staying hungry for a month was not easy).

Mixer – If you don’t have one, get one!

Emergency – DON’T PANIC. Carry scissors all the time. If you are suffocating, then chop off the rubber-bands. (The doc never told me this).

Oral hygiene – Brush with a soft toothbrush. Rinse (with mouthwash) after every major liquid intake. Use lukewarm salt water.

Fighting Depression – The two days between the accident and surgery were the worst as I didn’t know the future of my jaws. After the surgery it was a relief to know there was not permanent damage but the first few days were disturbing. I frequently went to the introspect-mode and used to feel hopeless.

Elastic band – My rubber bands lost elasticity because of yawning, coughing and talking. After 2-3 weeks, the doc fixed more bands.

Internet – The internet is a very useful resource. Once you suffer from something, you read every relevant bit of information you can find and become an expert in it. I think I am an expert in mandible fractures. About fora (forums), there are people who suffered similar accidents and believe me, what you are suffering would most probably be someone else’s problem too. It feels good to find people who suffered similarly and with whom you can connect yourself. But beware, people have suffered worse problems so it might not be necessary to panic for every trauma someone else had.

Healing – It’s obvious that everyone’s healing process is different. It depends on your fracture type, the doctors’ treatment and your own physical health.

Complete recovery – A month of no brushing decays teeth, also the accident may have caused dental damage. So after removing the braces is a good time to see a dentist. I had my root canal done in India because dental care there is excellent (and of course, inexpensive). Since the jaw was shut for weeks, the mouth muscles have lost their elasticity and it takes months to open the jaw completely.

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Tailpiece – If you are suffering from a mandible fracture and wish to get in touch, just leave a comment and I shall be glad to offer any help. After all, a bro is what a bro does!

This post is for my friends (especially in India) who read about my accident from an earlier post and have been concerned about my well-being.

I had multiple fractures in my jaw and after a surgery, my mouth was wired shut for nearly six weeks. For six weeks, life was pathetic. I couldn’t eat solid food (hence the post on liquid diet) and lost lots of kilos. A stifled cough, no yawning and a mouth-explosion every time I sneezed — it was a miserable experience and I hope you never have to go through it. Just yesterday, the agony was put an end to, as the doctor removed the rubber-bands.

I can now open my mouth albeit partially (a finger-width) but enough to pass a toothbrush through. Brushing and being able to talk after six weeks was quite relieving. The surgery to remove braces will take place next week. My wrist injury is also slowly healing. The total healing of the mouth will take a couple of months. Hanging in there,

Abhishek

Before going to the doctor to get the rubber-bands removed

In this post, I will show how algebraic curves and algebraic numbers are related. I shall continue using the notations developed in the previous post here. You may read that post to refresh the theory of algebraic curves.

The unifying feature between the two is the concerned rings are Dedekind domains. A Dedekind domain is an integrally closed Noetherian domain of dimension 1. It has two excellent properties,

• The localization at every prime ideal is a dvr.
• The integral closure of a DD in a finite separable extension of its field of fractions is again a DD.

In the curves case, suppose we have two curves (i.e., non-singular projective varieties of dimension 1) and a non-constant morphism between them.

$\displaystyle \phi : C_1 \rightarrow C_2.$

Then as seen earlier, it induces an inclusion of function fields,

$\displaystyle \phi^* : k(C_2) \hookrightarrow k(C_1).$

#### Ramification

We shall now define what ramification at a point ${P \in C_1}$ means. Intuitively, it means there is a knot at ${P}$. Let us use some commutative algebra to make this notion precise.

Let ${Q = \phi(P)}$. Let ${t}$ be an element of the function field ${k(C_2)}$ that generates the maximal ideal of the local ring (also a dvr) ${k[C_2]_Q}$. Let ${e_P}$ be the ${P}$-order of ${\phi^*(t) = t \circ \phi}$ as a function of ${k(C_1)}$. Then ${\phi}$ is ramified at ${P}$ if ${e_P>1}$ and unramified if ${e_P=1}$ for every point ${P\in C_1}$.

We have the following :

#### Theorem:

Let ${\phi : C_1 \rightarrow C_2}$ be a non-constant map of curves. Then,

• For every point ${Q\in C_2}$,

$\displaystyle \text{deg } \phi = \displaystyle\sum_{P\mapsto Q} e_P.$

• For all but finitely many points ${Q\in C_2}$,

$\displaystyle |\phi^{-1}(Q)| = \text{deg}_s \phi,$

where ${\text{deg}_s}$ denotes the degree of separability of ${k(C_1) / \phi^* (k(C_2))}$.

• If ${\psi : C_2 \rightarrow C_3}$ is another non-constant map of curves, then

$\displaystyle e_{\psi\circ\phi}(P) = e_\phi (P) . e_\psi (\phi P).$

#### Corresponding results in number theory

The first result corresponds to the identity ${\displaystyle\sum_{i=1}^g e_i f_i = [L:K]}$ for number fields ${L/K}$. The second one says that only finitely many primes ramify and the third result is the multiplicativity of ramification indices in a tower of number fields. Let us state these results more precisely in the following

#### Theorem:

Let ${K \subseteq L}$ be number fields with ${[L:K]< \infty}$. Let ${\mathcal O_K}$ and ${\mathcal O_L}$ be the corresponding rings of integers. Then,

• For every prime ${P}$ in ${\mathcal O_K}$, we have

$\displaystyle P \mathcal O_L = Q_1^{e_1} Q_2^{e_2} \cdots Q_g^{e_g}$

where the ${Q_i}$‘s are primes in ${\mathcal O_L}$. Then, $\displaystyle \displaystyle\sum_{i=1}^g e_i f_i = [L:K]$ holds, where ${f_i}$ is the inertial degree given by ${[ \mathcal O_L/Q_i : \mathcal O_K/P ]}$.

• At most finitely many primes of ${\mathcal O_K}$ ramify in ${\mathcal O_L}$. (A prime ${P}$ of ${\mathcal O_K}$ is said to ramify in ${L}$ if ${e_i>1}$ for some ${i}$).
• If ${M}$ is a number field containing ${L}$, then for every prime ${P}$ of ${\mathcal O_K}$,

$\displaystyle e_{M/K} = e_{M/L}. e_{L/K}.$

#### More analogy

The similarity between number fields and algebraic curves does not end here. In the number theoretic case, we have the class group of a number field which is the quotient of the free abelian group on prime ideals modulo the principle ideals. Similarly, for algebraic curves we have the Picard group which is the free abelian group on divisors modulo principal divisors. Both groups turn to be finite (after some struggle in proving it).

Finally, the analog of the exact sequence in number theory (here ${U_K}$ is the group of units)

$\displaystyle 1 \rightarrow U_K \rightarrow K^* \rightarrow \displaystyle \begin{pmatrix} \text{fractional} \\ \text{ideals of }\mathcal O_K \end{pmatrix} \rightarrow C_K \rightarrow 0$

is the exact sequence of degree-zero divisors

$\displaystyle 1 \rightarrow K^* \rightarrow K(C)^* \rightarrow \text{Div}^0(C) \rightarrow \text{Pic}^0(C) \rightarrow 0.$

They were the brilliant schemes of Grothendieck and his co-workers that unified algebraic geometry and number theory with tools (results) from the former being made available to the latter. He was able to prove the Weil conjectures with these abstract unified objects known as schemes but more on that later (after I study it!)

In this post, we shall see some results about algebraic curves. For the past week, I have been studying algebraic curves and the great Riemann-Roch theorem. I won’t go in the details but will only sketch the basic theory behind their study.

#### Algebraic curves

By a curve we will mean a projective variety of dimension 1. For the sake of simplicity let us assume the base field ${k}$ to be algebraically closed. (This is far from sufficient; indeed the most interesting applications in number theory will have ${k}$ as the rational numbers or finite fields or ${p}$-adic fields). One can think of the curve as the locus of the zero set of an irreducible polynomial ${F(X,Y) \in k[X,Y]}$. This as it is, is an affine curve and we attach points of infinity as necessary by homogenizing ${F}$. Note that it is not necessary that the curve be generated by just one polynomial. It may be possible that it is the intersection of higher dimensional varieties in projective space of higher dimension. The only requirement is that the dimension of this variety be 1.

#### Function field of a curve

Given a curve ${C}$, we can associate a field to it, namely the function field of ${C}$. As a concrete example, let ${C}$ be the projective circle given by the homogeneous equation ${F(X,Y,Z) = X^2 + Y^2 - Z^2}$. (If you are not habituated to using projective coordinates, just substitute 1 for ${Z}$ and everything should work fine). Then the function field of ${C}$ is

$\displaystyle k(C) = \text{Field of fractions of } \displaystyle\frac{k[X,Y,Z]}{(X^2+Y^2-Z^2)}.$

Note that asking the dimension of ${C}$ to be 1 is the same as ${k9C)}$ having a transcendence degree 1 over ${k}$.

Corresponding to any point ${P=[a:b:c]}$ on the curve, there is a discrete valuation ring ${k[C]_P}$ which is the localization of the domain ${\displaystyle\frac{k[X,Y,Z]}{(X^2+Y^2-Z^2)}}$ at the maximal ideal ${(X-a, Y-b, Z-c)}$. A fundamental theorem in algebraic geometry says that the point ${P}$ on ${C}$ (in fact any variety) is non-singular if and only if ${k[C]_P}$ is a regular local ring.

#### Maps between curves

By a map between curves ${C_1}$ and ${C_2}$ we will mean a morphism of the corresponding projective varieties. If

$\displaystyle \phi : C_1 \rightarrow C_2$

is a morphism of curves, then ${\phi}$ is either constant or surjective! Also, ${\phi}$ induces a map between the function fields viz.

$\displaystyle \phi^* : k(C_2) \hookrightarrow k(C_1) \qquad \phi^*(f) = f \circ \phi.$

If ${\phi}$ is nonconstant this gives a finite extension of fields, ${[k(C_1) : \phi^*(k(C_2))]}$ and we define the degree ${deg(\phi)}$ to be the degree of this extension. A map of degree 1 is an isomorphism.

#### Categorical equivalence between curves and function fields

We saw that a curve ${C}$ gives a function field ${k(C)}$ of transcendence degree 1 over ${k}$. Morphisms of curves give an inclusion of function fields. Indeed, this functoriality goes beyond, it’s a categorical equivalence. Given a field ${\mathbb K/k}$ of transcendence degree 1, one proves that the collection of local rings ${R}$ such that ${k \subset R \subset \mathbb K}$ actually define a non-singular projective curve. (All the ${R}$‘s will be dvrs since tr. deg${(\mathbb K/k)=1}$). The morphisms in this category are inclusion maps of fields and they give morphisms of curves.

If you have read this far, then you are very close to understanding the connection between algebraic curves and algebraic number fields. This is explained in the next post here.

#### References

• R. Hartshorne, Algebraic Geometry (Chapter 1)
• J. Silverman, Arithmetic of Elliptic Curves (Chapter 2)
(You might have read about my running the first $\pi$ miles. This is a rather different experience).

(Edit: 6 July, ’12 — If you have fractured a jaw and are looking for a liquid diet, see this post I wrote recently).

Today evening I completed running $\pi$ miles on the Wabash river trail. I have done it more than a hundred times in the last few months. But today was special. Completing the track was a liberating moment. For one, I completed it with less-than-usual oxygen; I breathed only through my nose and not once through the mouth.

As I write this, I feel the steel wires and rubber-bands in my mouth. My teeth are fixed, the jaws are immovable and there are six stitches on the chin. I can feel one molar chipped off and one week after the accident, I feel pain in my wrist.

It was on the evening of 26th May that the four of us set out for biking. We were around 10 miles from Purdue where I had this accident. It was a reckless daredevilry, something that I am not proud of & something I wouldn’t like to recollect¹. After the accident, we called a friend and also 911 (the latter came before) and I was taken to the Emergency room of the hospital². I had the X-ray and CT-scan done but already knew my jaw was fractured. The doctor stitched the chin-injury and discharged me with the desperately-needed painkillers. Three days later, I had a surgery with an oral surgeon who (after knocking me unconscious) implanted steel braces and rubber-bands in my mouth thus completely immobilizing any jaw movement. For six weeks, I shall not be able to open my mouth. No coughing or yawning please! And only liquid diet.

#### Tailpiece:

I realize that the pain-killers have subdued any painful sensation in my jaws. Most of me is perfectly fine, most importantly my brain. And I am doing Mathematics with acetaminophen. Just like Erdös!³ Hope I am able to come up with some theorem in these six weeks!

##### ³ If you are unaware, the legendary mathematician Paul Erdös took amphetamines and said the drug helped him do mathematics. Read his biography here.

Abhishek Parab

I? An Indian. A mathematics student. A former engineer. A rubik's cube addict. A nature photographer. A Pink Floyd fan. An ardent lover of Chess & Counter-Strike.

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### Quotable Quotes

ABHISHEK PARAB
“Do not think; let the equation think for you”

PAUL HALMOS
”You cannot be perfect, but if you won’t try, you won’t be good enough”

ALBERT EINSTEIN
“Don’t worry about your maths problems; I assure you, mine are greater”

THE BEST MATH JOKE
"A comathematician is a device for turning cotheorems into ffee"

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