# Conference Amplitudes 2015 – Air on the Superstring

One of the first pieces of Bach ever recorded was August Wilhelmj’s arrangement of the Orchestral Suite in D major. Today the transcription for violin and piano goes by the moniker Air on the G String. It’s an inspirational and popular work in all it’s many incarnations, not least this one featuring my favourite cellist Yo-Yo Ma.

This morning we heard the physics version of Bach’s masterpiece. Superstrings are nothing new, of course. But recently they’ve received a reboot courtesy of Dr. David Skinner among others. The ambitwistor string is an infinite tension version which only admit right-moving vibrations! At first the formalism looks a little daunting, until you realise that many calculations follow the well-trodden path of the superstring.

Now superstring amplitudes are quite difficult to compute. So hard, in fact, that Dr. Oliver Schloterrer devoted an entire talk to understanding particular functions that emerge when scattering just  $4$ strings at next-to-leading order. Mercifully, the ambitwistor string is far more well-behaved. The resulting amplitudes are rather beautiful and simple. To some extent, you trade off the geometrical aesthetics of the superstring for the algebraic compactness emerging from the ambitwistor approach.

This isn’t the first time that twistors and strings have been combined to produce quantum field theory. The first attempt dates back to 2003 and work of Edward Witten (of course). Although hugely influential, Witten’s theory was esoteric to say the least! In particular nobody knows how to encode quantum corrections in Witten’s language.

Ambitwistor strings have no such issues! Adding a quantum correction is easy – just put your theory on a donut. But this conceptually simple step threatened a roadblock for the research. Trouble was, nobody actually knew how to evaluate the resulting formulae.

Nobody, that was, until last week! Talented folk at Oxford and Cambridge managed to reduce the donutty problem to the original spherical case. This is an impressive feat – nobody much suspected that quantum corrections would be as easy as a classical computation!

There’s a great deal of hope that this idea can be rigorously extended to higher loops and perhaps even break the deadlock on maximal supergravity calculations at $7$-loop level. The resulting concept of off-shell scattering equations piqued my interest – I’ve set myself a challenge to use them in the next 12 months!

Scattering equations, you say? What are these beasts? For that we need to take a closer look at the form of the ambitwistor string amplitude. It turns out to be a sum over the solutions of the following equations

$\sum_{i\neq j}\frac{s_{ij}}{z_i - z_j}=0$

The $s_{ij}$ are just two particle invariants – encoding things you can measure about the speed and angle of particle scattering. And the $z_i$ are just some bonus variables. You’d never dream of introducing them unless somebody told you to! But yet they’re exactly what’s required for a truly elegant description.

And these scattering equations don’t just crop up in one special theory. Like spies in a Cold War era film, they seem to be everywhere! Dr. Freddy Cachazo alerted us to this surprising fact in a wonderfully engaging talk. We all had a chance to play detective and identify bits of physics from telltale clues! By the end we’d built up an impressive spider’s web of connections, held together by the scattering equations.

Freddy’s talk put me in mind of an interesting leadership concept espoused by the conductor Itay Talgam. Away from his musical responsibilities he’s carved out a niche as a business consultant, teaching politicians, researchers, generals and managers how to elicit maximal productivity and creativity from their colleagues and subordinates. Critical to his philosophy is the concept of keynote listening – sharing ideas in a way that maximises the response of your audience. This elusive quality pervaded Freddy’s presentation.

Following this masterclass was no mean feat, but one amply performed by my colleague Brenda Penante. We were transported to the world of on-shell diagrams – a modern alternative to Feynman’s ubiquitous approach. These diagrams are known to produce the integrand in planar $\mathcal{N}=4$ super-Yang-Mills theory to all orders! What’s more, the answer comes out in an attractive $d \log$ form, ripe for integration to multiple polylogarithms.

Cunningly, I snuck the word planar into the paragraph above. This approximation means that the diagrams can be drawn on a sheet of paper rather than requiring $3$ dimensions. For technical reasons this is equivalent to working in the theory with an infinite number of color charges, not just the usual $3$ we find for the strong force.

Obviously, it would be helpful to move beyond this limit. Brenda explained a decisive step in this direction, providing a mechanism for computing all leading singularities of non-planar amplitudes. By examining specific examples the collaboration uncovered new structure invisible in the planar case.

Technically, they observed that the boundary operation on a reduced graph identified non-trivial singularities which can’t be understood as the vanishing of minors. At present, there’s no proven geometrical picture of these new relations. Amazingly they might emerge from a 1,700-year-old theorem of Pappus!

Bootstraps were back on the agenda to close the session. Dr. Agnese Bissi is a world-expert on conformal field theories. These models have no sense of distance and only know about angles. Not particularly useful, you might think! But they crop up surprisingly often as approximations to realistic physics, both in particle smashing and modelling materials.

Agnese took a refreshingly rigorous approach, walking us through her proof of the reciprocity principle. Until recently this vital tool was little more than an ad hoc assumption, albeit backed up by considerable evidence. Now Agnese has placed it on firmer ground. From here she was able to “soup up” the method. The supercharged variant can compute OPE coefficients as well as dimensions.

Alas, it’s already time for the conference dinner and I haven’t mentioned Dr. Christian Bogner‘s excellent work on the sunrise integral. This charmingly named function is the simplest case where hyperlogarithms are not enough to write down the answer. But don’t just take it from me! You can now hear him deliver his talk by visiting the conference website.

Conversations

I’m very pleased to have chatted with Professor Rutger Boels (on the Lagrangian origin of Yang-Mills soft theorems and concerning the universality of subheading collinear behaviour) and Tim Olson (about determining the relative sign between on-shell diagrams to ensure cancellation of spurious poles).

Note: this post was originally written on Thursday 9th July but remained unpublished. I blame the magnificent food, wine and bonhomie at the conference dinner!

# Financial Times Christmas Carol!

Away from my physics life I spend a lot of time singing. About 6 months ago I co-founded a choir. The development of the group has been phenomenally rapid, culminating in recording a new Christmas carol for the Financial Times, composed by the acclaimed baritone Roderick Williams. Do have a listen and let me know what you think!

I see mathematics and music as natural intellectual cousins. Both involve artistic creativity within certain constraints. As a researcher, I must learn the subject and find new concepts. As a musician I’ll certainly study the notes, but it’s that spark of original interpretation which brings the music alive.

If you liked the FT carol, have a listen to our other recent recording below!

I’ll start this post by tying up some loose ends from last time. Before we get going there’s no better recommendation for uplifting listening than this marvellous recording. Hopefully it’ll help motivate and inspire you (and I) as we journey deeper into the weird and wonderful world of algebra and geometry.

I promised a proof that for algebraically closed fields $k$ every Zariski open set is dense in the Zariski topology. Quite a mouthful at this stage of a post, I admit. Basically what I’m showing is that Zariski open sets are really damn big, only in a mathematically precise way. But what of this ‘algebraically closed’ nonsense? Time for a definition.

Definition 3.1 A field $k$ is algebraically closed if every nonconstant polynomial in $k[x]$ has a root in $k$.

Let’s look at a few examples. Certainly $\mathbb{R}$ isn’t algebraically closed. Indeed the polynomial $x^2 + 1$ has no root in $\mathbb{R}$. By contrast $\mathbb{C}$ is algebraically closed, by virtue of the Fundamental Theorem of Algebra. Clearly no finite field is algebraically closed. Indeed suppose $k=\{p_1,\dots ,p_n\}$ then $(x-p_1)\dots (x-p_n) +1$ has no root in $k$. We’ll take a short detour to exhibit another large class of algebraically closed fields.

Definition 3.2 Let $k,\ l$ be fields with $k\subset l$. We say that $l$ is a field extension of $k$ and write $l/k$ for this situation. If every element of $l$ is the root of a polynomial in $k[x]$ we call $l/k$ an algebraic extension. Finally we say that the algebraic closure of $k$ is the algebraic extension $\bar{k}$ of $k$ which is itself algebraically closed.

(For those with a more technical background, recall that the algebraic closure is unique up to $k$-isomorphisms, provided one is willing to apply Zorn’s Lemma).

The idea of algebraic closure gives us a pleasant way to construct algebraically closed fields. However it gives us little intuition about what these fields ‘look like’. An illustrative example is provided by the algebraic closure of the finite field of order $p^d$ for $p$ prime. We’ll write $\mathbb{F}_{p^d}$ for this field, as is common practice. It’s not too hard to prove the following

Theorem 3.3 $\mathbb{F}_{p^d}=\bigcup_{n=1}^{\infty}\mathbb{F}_{p^{n!}}$

Proof Read this PlanetMath article for details.

Now we’ve got a little bit of an idea what algebraically closed fields might look like! In particular we’ve constructed such fields with characteristic $p$ for all $p$. From now on we shall boldly assume that for our purposes

every field $k$ is algebraically closed

I imagine that you may have an immediate objection. After all, I’ve been recommending that you use $\mathbb{R}^n$ to gain an intuition about $\mathbb{A}^n$. But we’ve just seen that $\mathbb{R}$ is not algebraically closed. Seems like we have an issue.

At this point I have to wave my hands a bit. Since $\mathbb{R}^n$ is a subset of $\mathbb{C}^n$ we can recover many (all?) of the geometrical properties we want to study in $\mathbb{R}^n$ by examining them in $\mathbb{C}^n$ and projecting appropriately. Moreover since $\mathbb{C}^n$ can be identified with $\mathbb{R}^{2n}$ in the Euclidean topology, our knowledge of $\mathbb{R}^n$ is still a useful intuitive guide.

However we should be aware that when we are examining affine plane curves with $k=\mathbb{C}$ they are in some sense $4$ dimensional objects – subsets of $\mathbb{C}^2$. If you can imagine $4$ dimensional space then you are a better person than I! That’s not to say that these basic varieties are completely intractable though. By looking at projections in $\mathbb{R}^3$ and $\mathbb{R}^2$ we can gain a pretty complete geometric insight. And this will soon be complemented by our burgeoning algebraic understanding.

Now that I’ve finished rambling, here’s the promised proof!

Lemma 3.4 Every nonempty Zariski open subset of $\mathbb{A}^1$ is dense.

Proof Recall that $k[x]$ is a principal ideal domain. Thus any ideal $I\subset k[x]$ may be written $I=(f)$. But $k$ algebraically closed so $f$ splits into linear factors. In other words $I = ((x-a_1)\dots (x-a_n))$. Hence the nontrivial Zariski closed subsets of $\mathbb{A}^1$ are finite, so certainly the Zariski open subsets of $\mathbb{A}^1$ are dense. $\blacksquare$

I believe that the general case is true for the complement of an irreducible variety, a concept which will be introduced next. However I haven’t been able to find a proof, so have asked here.

How do varieties split apart? This is a perfectly natural question. Indeed many objects, both in mathematics and the everyday world, are made of some fundamental building block. Understanding this ‘irreducible atom’  gives us an insight into the properties of the object itself. We’ll thus need a notion for what constitutes an ‘irreducible’ or ‘atomic’ variety.

Definition 3.5 An affine variety $X$ is called reducible if one can write $X=Y\cup Z$ with $Y,\ Z$ proper subsets of $X$. If $X$ is not reducible, we call it irreducible.

This seems like a good and intuitive way of defining irreducibility. But we don’t yet know that every variety can be constructed from irreducible building blocks. We’ll use the next few minutes to pursue such a theorem.

As an aside, I’d better tell you about some notational confusion that sometimes creeps in. Some authors use the term algebraic set for  my term affine variety. Such books will often use the term affine variety to mean irreducible algebraic set. For the time being I’ll stick to my guns, and use the word irreducible when it’s necessary!

Before we go theorem hunting, let’s get an idea about what irreducible varieties look like by examining some examples. The ‘preschool’ example is that $V(x_1 x_2)\subset \mathbb{A}^2$ is reducible, for indeed $V(x_1 x_2) = V(x_1)\cup V(x_2)$. This is neither very interesting nor really very informative, however.

A better example is the fact that $\mathbb{A}^1$ is irreducible. To see this, recall that earlier we found that the only proper subvarieties of $\mathbb{A}^1$ are finite. But $k$ is algebraically closed, so infinite. Hence we cannot write $\mathbb{A}^1$ as the union of two proper subvarities!

What about the obvious generalization of this to $\mathbb{A}^n$? Turns out that it is indeed true, as we might expect. For the sake of formality I’ll write it up as a lemma.

Lemma 3.6 $\mathbb{A}^n$ is irreducible

Proof Suppose we could write $\mathbb{A}^n=V(f)\cup V(g)$. By Lemma 2.5 we know that $V(f)\cup V(g) = V((f)\cap (g))$. But $(f)\cap(g)\supset (fg)$ so $V((f)\cap(g))\subset V(fg)$ again by Lemma 2.5. Conversely if $x\in V(fg)$ then either $f(x) = 0$ or $g(x) = 0$, so $x \in V(f)\cup V(g)$. This shows that $V(f)\cup V(g)=V(fg)$.

Now $V(fg)=\mathbb{A}^n$ immediately tells us $fg(x) = 0 \ \forall x\in k$. Suppose that $f$ is nonzero. We’ll prove that $g$ is the zero polynomial by induction on $n$. Then $V(g)=\mathbb{A}^n$ so $\mathbb{A}^n$ not irreducible, as required.

We first note that since $k$ algebraically closed $k$ infinite. For $n=1$ suppose $f,\ g \neq 0$. Then $f,\ g$ are each zero at finite sets of points. Thus since $k$ infinite, $fg$ is not the zero polynomial, a contradiction.

Now let $n>1$.  Consider $f,\ g$ nonzero polynomials in $k[\mathbb{A}^n]$. Fix $x_n \in k$. Then $f,\ g$ polynomials in $k[\mathbb{A}^{n-1}]$. For some $x_n$, $f,\ g$ nonzero as polynomials in $k[\mathbb{A}^{n-1}]$. By the induction hypothesis $fg\neq 0$. This completes the induction. $\blacksquare$

I’ll quickly demonstrate that $\mathbb{A}^n$ is quite strange, when considered as a topological space with the Zariski topology! Indeed let $U$ and $V$ be two nonempty open subsets. Then $U\cap V\neq \emptyset$. Otherwise $\mathbb{A}^n\setminus U,\ \mathbb{A}^n\setminus V$ would be proper closed subsets (affine subvarieties) which covered $\mathbb{A}^n$, violating irreducibility. This is very much not what happens in the Euclidean topology! Similarly we now have a rigorous proof that an open subset $U$ of $\mathbb{A}^n$ is dense. Otherwise $\bar{U}$ and $\mathbb{A}^n\setminus U$ would be proper subvarieties covering $\mathbb{A}^n$.

It’s all very well looking for direct examples of irreducible varieties, but in doing so we’ve forgotten about algebra! In fact algebra gives us a big helping hand, as the following theorem shows. For completeness we first recall the definition of a prime ideal.

Definition 3.7 $\mathfrak{p}$ is a prime ideal in $R$ iff whenever $fg \in \mathfrak{p}$ we have $f\in \mathfrak{p}$ or $g \in \mathfrak{p}$. Equivalently $\mathfrak{p}$ is prime iff $R/\mathfrak{p}$ is an integral domain.

Theorem 3.8 Let $X$ be a nonempty affine variety. Then $X$ irreducible iff $I(X)$ a prime ideal.

Proof [“$\Rightarrow$“] Suppose $I(X)$ not prime. Then $\exists f,g \in k[\mathbb{A}^n]$ with $fg \in I(X)$ but $f,\ g \notin I(X)$. Let $J_1 = (I(X),f)$ and $J_2 = (I(X),g)$. Further define $X_1 = V(J_1), \ X_2 = V(J_2)$. Then $V(X_1), \ V(X_2) \subset X$ so proper subsets of $\mathbb{A}^n$. On the other hand $X\subset X_1 \cup X_2$. Indeed if $P\in X$ then $fg(P)=0$ so $f(P)=0$ or $g(P)=0$ so $P \in X_1\cup X_2$.

[“$\Leftarrow$“] Suppose $X$ is reducible, that is $\exists X_1,\ X_2$ proper subvarieties of $X$ with $X=X_1\cup X_2$. Since $X_1$ a proper subvariety of $X$ there must exist some element $f \in I(X_1)\setminus I(X)$. Similarly we find $g\in I(X_2)\setminus I(X)$. Hence $fg(P) = 0$ for all $P$ in $X_1\cup X_2 = X$, so certainly $fg \in I(X)$. But this means that $I(X)$ is not prime. $\blacksquare$

This easy theorem is our first real taste of the power that abstract algebra lends to the study of geometry. Let’s see it in action.

Recall that a nonzero principal ideal of the ring $k[\mathbb{A}^n]$ is prime iff it is generated by an irreducible polynomial. This is an easy consequence of the fact that $k[\mathbb{A}^n]$ is a UFD. Indeed a nonzero principal ideal is prime iff it is generated by a prime element. But in a UFD every prime is irreducible, and every irreducible is prime!

Using the theorem we can say that every irreducible polynomial $f$ gives rise to an irreducible affine hypersurface $X$ s.t. $I(X)=(f)$. Note that we cannot get a converse to this – there’s nothing to say that $I(X)$ must be principal in general.

Does this generalise to ideals generated by several irreducible polynomials? We quickly see the answer is no. Indeed take $f = x\, g = x^2 + y^2 -1$ in $k[\mathbb{A}^2]$. These are both clearly irreducible, but $(f,g)$ is not prime. We can see this in two ways. Algebraically $y^2 \in (f,g)$ but $y \notin (f,g)$. Geometrically, recall Lemma 2.5 (3). Also note that by definition $(f,g) = (f)+(g)$. Hence $V(f,g) = V(f)\cap V(g)$. But $V(f) \cap V(g)$ is clearly just two distinct points (the intersection of the line with the circle). Hence it is reducible, and by our theorem $(f,g)$ cannot be prime.

We can also use the theorem to exhibit a more informative example of a reducible variety. Consider $X = V(X^2Y - Y^2)$. Clearly $\mathfrak{a}=(X^2Y-Y^2)$ is not prime for $Y(X^2 - Y) \in \mathfrak{a}$ but $Y\notin \mathfrak{a}, \ X^2 - Y \notin \mathfrak{a}$. Noting that $\mathfrak{a}=(X^2-Y)\cap Y$ we see that geometrically $X$ is the union of the $X$-axis and the parabola $Y=X^2$, by Lemma 2.5.

Having had such success with prime ideals and irreducible varieties, we might think – what about maximal ideals? Turns out that they have a role to play too. Note that maximal ideals are automatically prime, so any varieties they generate will certainly be irreducible.

Definition 3.9 An ideal $\mathfrak{m}$ of $R$ is said to be maximal if whenever $\mathfrak{m}\subset\mathfrak{a}\subset R$ either $\mathfrak{a} = \mathfrak{m}$ or $\mathfrak{a} = R$. Equivalently $\mathfrak{m}$ is maximal iff $R/\mathfrak{m}$ is a field.

Theorem 3.10 An affine variety $X$ in $\mathbb{A}^n$ is a point iff $I(X)$ is a maximal ideal.

Proof  [“$\Rightarrow$“] Let $X = \{(a_1, \dots , a_n)\}$ be a single point. Then clearly $I(X) = (X_1-a_1,\dots ,X_n-a_n)$. But $k[\mathbb{A}^n]/I(X)$ a field. Indeed $k[\mathbb{A}^n]/I(X)$ isomorphic to $k$ itself, via the isomorphism $X_i \mapsto a_i$. Hence $I(X)$ maximal.

[“$\Leftarrow$“] We’ll see this next time. In fact all we need to show is that $(X_1-a_1,\dots,X_n-a_n)$ are the only maximal ideals. $\blacksquare$

Theorems 3.8 and 3.10 are a promising start to our search for a dictionary between algebra and geometry. But they are unsatisfying in two ways. Firstly they tell us nothing about the behaviour of reducible affine varieties – a very large class! Secondly it is not obvious how to use 3.8 to construct irreducibly varieties in general. Indeed there is an inherent asymmetry in our knowledge at present, as I shall now demonstrate.

Given an irreducible variety $X$ we can construct it’s ideal $I(X)$ and be sure it is prime, by Theorem 3.8. Moreover we know by Lemma 2.5 that $V(I(X))=X$, a pleasing correspondence. However, given a prime ideal $J$ we cannot immediately say that $V(J)$ is prime. For in Lemma 2.5 there was nothing to say that $I(V(J))=J$, so Theorem 3.8 is useless. We clearly need to find a set of ideals for which $I(V(J))=J$ holds, and hope that prime ideals are a subset of this.

It turns out that such a condition is satisfied by a class called radical ideals. Next time we shall prove this, and demonstrate that radical ideals correspond exactly to algebraic varieties. This will provide us with the basic dictionary of algebraic geometry, allowing us to proceed to deeper results. The remainder of this post shall be devoted to radical ideals, and the promised proof of an irreducible decomposition.

Definition 3.11 Let $J$ be an ideal in a ring $R$. We define the radical of $J$ to be the ideal $\sqrt{J}=\{f\in R : f^m\in J \ \textrm{some} \ m\in \mathbb{N}\}$. We say that $J$ is a radical ideal if $J=\sqrt{J}$.

(That $\sqrt{J}$ is a genuine ideal needs proof, but this is merely a trivial check of the axioms).

At first glance this appears to be a somewhat arbitrary definition, though the nomenclature should seem sensible enough. To get a more rounded perspective let’s introduce some other concepts that will become important later.

Definition 3.12polynomial function or regular function on an affine variety $X$ is a map $X\rightarrow k$ which is defined by the restriction of a polynomial in $k[\mathbb{A}^n]$ to $X$. More explicitly it is a map $f:X\rightarrow k$ with $f(P)=F(P)$ for all $P\in X$ where $F\in k[\mathbb{A}^n]$ some polynomial.

These are eminently reasonable quantities to be interested in. In many ways they are the most obvious functions to define on affine varieties. Regular functions are the analogues of smooth functions in differential geometry, or continuous functions in topology. They are the canonical maps.

It is obvious that a regular function $f$ cannot in general uniquely define the polynomial $F$ giving rise to it. In fact suppose $f(P)=F(P)=G(P) \ \forall P \in X$. Then $F-G = 0$ on $X$ so $F-G\in I(X)$. This simple observation explains the implicit claim in the following definition.

Definition 3.13 Let $X$ be an affine variety. The coordinate ring $k[X]$ is the ring $k[\mathbb{A}^n]|_X=k[\mathbb{A}^n]/I(X)$. In other words the coordinate ring is the ring of all regular functions on $X$.

This definition should also appear logical. Indeed we define the space of continuous functions in topology and the space of smooth functions in differential geometry. The coordinate ring is merely the same notion in algebraic geometry.  The name  ‘coordinate ring’ arises since clearly $k[X]$ is generated by the coordinate functions $x_1,\dots ,x_n$ restricted to $X$. The reason for our notation $k[x_1,\dots ,x_n]=k[\mathbb{A}^n]$ should now be obvious. Note that the coordinate ring is trivially a finitely generated $k$-algebra.

The coordinate ring might seem a little useless at present. We’ll see in a later post that it has a vital role in allowing us to apply our dictionary of algebra and geometry to subvarieties. To avoid confusion we’ll stick to $k[\mathbb{A}^n]$ for the near future. The reason for introducing coordinate rings was to link them to radical ideals. We’ll do this via two final definitions.

Definition 3.14 An element $x$ of a ring $R$ is called nilpotent if $\exists$ some positive integer $n$ s.t. $x^n=0$.

Definition 3.15 A ring $R$ is reduced if $0$ is its only nilpotent element.

Lemma 3.16 $R/I$ is reduced iff $I$ is radical.

Proof Let $x+I$ be a nilpotent element of $R/I$ i.e. $(x^n + I) = 0$. Hence $x^n \in I$ so by definition $x\in \sqrt{I}=I$. Conversely let $x\in R s.t. x^m \in I$. Then $x^m + I = 0$ in $R/I$ so $x+I = 0+I$ i.e. $x \in I$. $/blacksquare$

Putting this all together we immediately see that the coordinate ring $k[X]$ is a reduced, finitely generated $k$-algebra. That is, provided we assume that for an affine variety $X$, $I(X)$ is radical, which we’ll prove next time. It’s useful to quickly see that these properties characterise coordinate rings of varieties. In fact given any reduced, finitely generated $k$-algebra $A$ we can construct a variety $X$ with $k[X]=A$ as follows.

Write $A=k[a_1,\dots ,a_n]$ and define a surjective homomorphism $\pi:k[\mathbb{A}^n]\rightarrow A, \ x_i\mapsto a_i$. Let $I=\textrm{ker}(\pi)$ and $X=V(I)$. By the isomorphism theorem $A = k[\mathbb{A}^n]/I$ so $I$ is radical since $A$ reduced. But then by our theorem next time $X$ an affine variety, with coordinate ring $A$.

We’ve come a long way in this post, and congratulations if you’ve stayed with me through all of it! Let’s survey the landscape. In the background we have abstract algebra – systems of equations whose solutions we want to study. In the foreground are our geometrical ideas – affine varieties which represent solutions to the equations. These varieties are built out of irreducible blocks, like Lego. We can match up ideals and varieties according to various criteria. We can also study maps from geometrical varieties down to the ground field using the coordinate ring.

Before I go here’s the promised proof that irreducible varieties really are the building blocks we’ve been talking about.

Theorem 3.17 Every affine variety $X$ has a unique decomposition as $X_1\cup\dots\cup X_n$ up to ordering, where the $X_i$ are irreducible components and $X_i\not\subset X_j$ for $i\neq j$.

Proof (Existence) An affine variety $X$ is either irreducible or $X=Y\cup Z$ with $Y,Z$ proper subset of $X$. We similarly may decompose $Y$ and $Z$ if they are reducible, and so on. We claim that this process stops after finitely many steps. Suppose otherwise, then $X$ contains an infinite sequence of subvarieties $X\supsetneq X_1 \supsetneq X_2 \supsetneq \dots$. By Lemma 2.5 (5) & (7) we have $I(X)\subsetneq I(X_1) \subsetneq I(X_2) \subsetneq \dots$. But $k[\mathbb{A}^n]$ a Noetherian ring by Hilbert’s Basis Theorem, and this contradicts the ascending chain condition! To satisfy the $X_i \not\subset X_j$ condition we simply remove any such $X_i$ that exist in the decomposition we’ve found.

(Uniqueness) Suppose we have another decomposition $X=Y_1\cup Y_m$ with $Y_i\not\subset Y_j$ for $i\neq j$. Then $X_i = X_i\cap X = \bigcup_{j=1}^{m}( X_i\cap Y_j)$. Since $X_i$ is irreducible we must have $X_i\cap Y_j = X_i$ for some $j$. In particular $X_i \subset Y_j$. But now by doing the same with the $X$ and $Y$ reversed we find $X_k$ width $X_i \subset Y_j \subset X_k$. But this forces $i=k$ and $Y_j = X_i$. But $i$ was arbitrary, so we are done. $\blacksquare$

If you’re interested in calculating some specific examples of ideals and their associated varieties have a read about Groebner Bases. This will probably become a topic for a post at some point, loosely based on the ideas in Hassett’s excellent book. This question is also worth a skim.

I leave you with this enlightening MathOverflow discussion , tackling the irreducibility of polynomials in two variables. Although some of the material is a tad dense, it’s nevertheless interesting, and may be a useful future reference!

# Back From Holiday

refreshed, revived and ready to start this blog in earnest. Alas, no real post today! Nevertheless I have been working behind the scenes putting the finishing touches to the necessary background material. Okay, it is more than a little dry, and probably terrifying to the uninitiated. Don’t worry – I won’t need to use all of it right away. It should serve as a touchstone (for me as much as you) to ensure that I’m doing everything on a firm mathematical footing.

Talking of well written introductory books, I feel obliged to join the long list of individuals who have publicly praised Roger Penrose’s great work, The Road to Reality. I’ve owned the volume for several years, dipped into it now and then, but only over the past seven days have I truly appreciated its depth and scope. Certainly a worthwhile investment if you are interested in science at a more than superficial level!

Finally, if you find yourself at a loss for entertainment any time in the next few days I’d heartily endorse the fantastic Beethoven Prom Series currently in progress. I had been a little skeptical about this rather ‘obvious’ choice of repertoire, but Baremboim’s expert musicianship has won me over. Also they are really good pieces, after all.

I promise a proper post tomorrow – we’ll be talking about Affine Varieties. They’re not as scary as they sound, honest!

# A Slice of Algebra

and a nice cup of tea. I always find that helps. Before we get down to business, you might want to put this delightful recording on. It’s always nice to have a bit of background music, and Strauss just seems to fit with Algebra somehow.

A broad definition of Algebra could be the study of equations and their solutions. This is perhaps the type of algebra we’re all familiar with from school. Here’s a typical problem

Find $x\in \mathbb{R}$ given that $x^2-2x+1=0$

That was easy, of course. Let’s try another one

Find all $x,y \in \mathbb{R}$ such that $y-x^2 = 0$

Perhaps you had to think for a moment before realising that this just defines a parabola in 2D space, pictured below.

These example illustrate that the solutions to equations can come in the form of points, or curves, and it’s not hard to see that solutions to equations in sufficiently many variables can define surfaces of any dimension you like. For example the equation $z=0$ defines a plane in 3D space.

So we can easily see that Algebra gives rise to geometrical structures of the type we discussed in the last post. It should now seem natural to study geometrical structures from an algebraic point of view. Voila – we have the motivation for Algebraic Geometry.

There’s nothing to restrict us to studying the solutions (often referred to as zeroes) of a single equation. In fact many interesting and useful geometric constructions arise as the simultaneous zeroes of several equations. Can you see two equations in (some or all of) the variables $x,y,z$ whose simultaneous solutions give rise to the $y$-axis in 3D [2]?

The technical terminology for the collection of simultaneous zeroes of several equations is an algebraic set. It is the most fundamental object of study which we will focus on.

Here we reach a slight technical impasse. For what follows I’ll assume a familiarity with elementary abstract algebra as outlined on the Background page. This may be viewed as a technical toolkit for our forthcoming studies. I’ll also assume some very basic knowledge about Topology, though not much more than can be gleaned by a thorough reading of the Wikipedia page. If you’ve never come across abstract algebra before, now is the time to do some serious thinking! I can’t promise it’ll be easy, and it might take a couple of days to get your head around the concepts, but I promise you it’s worth it. I’ll be happy to answer any questions commented on the Background page, and will flesh out the currently sparse details in the near future.

Good luck!

[1] We only every consider polynomial equations, which are those of the form $f(x_1,\dots,x_n)=0$ where $f(x_1,\dots,x_n)$ is a finite sum of nonnegative integer powers  of products of the variables $x_1,\dots, x_n$. Thus $f(x,y)=x^2+y^2=0$, the circle, is admissible for study but $f(x,y)=x^y=0$ is not. It turns out that not much is lost by restricting our study to polynomials only. In some sense any mathematically interesting curve can be approximated arbitrarily closely by the set of solutions to polynomial equations. (This entirely depends on your definition of mathematically interesting though)!

[2] The equations are of course $x=0$ and $z=0$. Geometrically this is true since the $y$-axis is the intersection of the two planes defined by $x=0$ and $z=0$.

# So What Exactly Are We Doing Here?

Good afternoon. Over the next 12 weeks or so, this blog will grow into a collection of (mostly mathematical) ideas. If you’re at all interested in String Theory, Algebraic Geometry or Quantum Mechanics I should have something worthwhile to tell you. If you already don’t know what I’m talking about, don’t worry – I’ll attempt to make a great deal of what I write accessible to the diligent layman! I’ll start slowly and try not to lose people along the way. Hopefully this will end up being a cute introduction to a fascinating part of maths for people from all kinds of backgrounds.

The aim is to post about once a day, with the style being something between popular science and academic coursebook. I’ll try to tag posts accordingly, so it’s easy to tell what audience I’m pitching to. The first few days may be an extremely brief recap of some very foundational material to provide some explanation and background for non-mathematicians.

Occasionally I might discuss/opine/rant about other things, including music, sport, and just why we are getting quite so much rain. I hope this will provide a (necessary?) break from the maths. I’ll happily take requests for a post on a particular topic, but I can’t promise to become an instant expert.

Finally I can’t guarantee that everything I write will be entirely correct on first posting. Some of this material I’m learning for the first time myself, and it might take a couple of iterations before I fully grasp the concepts. If you think I’ve been unclear or don’t understand something, please do comment.

If you are still with me, well done! No more administrative faff, I promise! Have a couple of contrasting YouTube videos for your efforts, here and here.