A Strange Tensor Product

I'm taking a class on quantum computation at the moment, so I've been thinking a lot about tensor products (which I've written about before here). The tensor product is a weird operation. On vector spaces, it has a fairly straightforward definition, although it has some very strange, unintuitive properties. But the tensor product of general modules is even weirder.

Here's an example. What is the tensor product of $\mathbb{Z}/2\mathbb{Z}$ and $\mathbb{Z}/3\mathbb{Z}$ as $\mathbb{Z}$-modules? That is to say, what is $\mathbb{Z}/2\mathbb{Z} \otimes_{\mathbb{Z}} \mathbb{Z}/3\mathbb{Z}$? For convenience, I'll write $M$ for this tensor product from now on. Cyclic groups are not free $\mathbb{Z}$ modules, so we can't think about $M$ in the same way as we think about the tensor product of vector spaces. And the answer turns out to be very strange.

\[M = 0\]

This answer doesn't make very much sense to me. The tensor product of nonzero vector spaces $V \otimes W$ always contains copies of $V$ and $W$. How can two nonzero $\mathbb Z$-modules tensor to 0?

Although it is very counterintuitive, this fact is actually not very hard to prove. Suppose $a \otimes b \in M$. It's always true that $1 \cdot a \otimes b = a \otimes b$. But because $2$ and $3$ are relatively prime, we can express 1 as a linear combination of 2 and 3. In fact, $1 = 2 \cdot 2 - 3$. Then \[\begin{aligned} a \otimes b &= 1 \cdot (a \otimes b)\\ &= (2 \cdot 2 - 3) (a \otimes b)\\ &= 2 \cdot 2 (a \otimes b)- 3 (a \otimes b)\\ &= 2 [(2a) \otimes b] - a \otimes (3b)\\ &= 0 \end{aligned}\] Since we can use this trick to show that every element of $M$ is 0, $M$ must be the 0 module.

However, I don't find this proof very satisfying. To me, it doesn't feel like it explains why the tensor product should be 0. It just demonstrates that it is 0. Of course, the division between 'should' and 'is' is vague and subjective, but I would prefer a different perspective on the problem. Today, I came up with an alternative justification using tensor-hom adjunction. This justification isn't fully formal at the moment, but it gives me a sense of why I might think this tensor product should be 0.

Tensor-Hom Adjunction

To me, the tensor-hom adjunction is all about curried functions (which I've written about before here). The idea is essentially that if you have a bilinear function from $X \times Y$ to $Z$, you can express this as a linear function on $X$ that returns a linear function that takes in an element of $Y$ and returns an element of $Z$. This decomposition of a multi-argument function into single-variable functions that return other functions is precisely the idea of currying.

More formally, the tensor-hom adjunction gives us a natural isomorphism \[\text{Hom}(X \otimes Y, Z) \simeq \text{Hom}(X, \text{Hom}(Y, Z))\]

Using the Tensor-Hom Adjunction to Study Our Tensor Product

We can use this to study our module $M$. Let $Z$ be an arbitrary $\mathbb Z$-module. Then tensor-hom adjunction tells us that\[\text{Hom}(M, Z) = \text{Hom}(\mathbb Z /2 \mathbb Z \otimes \mathbb Z / 3 \mathbb Z, Z) \simeq \text{Hom}(\mathbb Z / 2 \mathbb Z, \text{Hom}(\mathbb Z / 3 \mathbb Z, Z))\]

The nonzero of $\mathbb Z / 3 \mathbb Z$ all have order 3. So their images under any $\mathbb Z$-linear map must have order 1 or order 3. Therefore, every element of $\text{Hom}(\mathbb Z / 3 \mathbb Z, Z)$ must have order 1 or 3 as an element of the group of $\mathbb Z$-linear maps. But the image of an element of $\mathbb Z / 2 \mathbb Z$ under a linear map must have order 1 or order 2. Since every element of the codomain has order 1 or order 3, we must map every element of $\mathbb Z / 2 \mathbb Z$ to the identity. So the only elements of $\text{Hom}(\mathbb Z / 2 \mathbb Z, \text{Hom}(\mathbb Z / 3 \mathbb Z, Z))$ are maps which send $\mathbb Z / 2 \mathbb Z$ to the zero map $\mathbb Z / 3 \mathbb Z \to Z$. This means that the only element of $\text{Hom}(M, Z)$ is the zero map, no matter what $Z$ is.

This is a pretty convincing argument to me that maybe $M$ should be 0. In particular, if $Z = M$, then we see that the only linear map from $M$ to itself is the zero map. So it looks like $M$ has to be 0. Furthermore, this argument generalizes naturally in the same way that the previous calculation does. The argument should work for $\mathbb Z / m \mathbb Z \otimes \mathbb Z / n \mathbb Z$ whenever $n$ and $m$ are relatively prime. So maybe it is a good perspective to keep in mind. We can always find nontrivial maps between nonzero vector spaces over a fixed field. But because modules can have restrictions on element orders, we can have modules which don't have nontrivial maps to each other. And this can result in their tensor products being 0.

I'm still not totally satisfied with my answer to this problem. The tensor product still seems to be a little spooky. If you have a different way of looking at the tensor product in general, or at this problem in particular, I'd love to hear it in the comments.

Today, I want to take Maxwell's equations and write them in the language of differential forms. The resulting equations are clearly covariant (i.e. they look the same after you apply a Lorentz transformation), and look a lot simpler than Maxwell's equations in vector notation. This is one of my favorite examples of how differential forms can make life easier.

Maxwell's Equations

In CGS units, Maxwell's equations are given by

1. $\nabla \cdot E = 4 \pi \rho$
2. $\nabla \cdot B = 0$
3. $\nabla \times E = -\frac 1 c \frac{\partial B}{\partial t}$
4. $\nabla \times B = \frac{4\pi}c J + \frac 1 c \frac{\partial E}{\partial t}$

$E$ is the electric field, $B$ is the magnetic field, $J$ is the electric current density, and $\rho$ is the electric charge density. $E, B$ and $J$ are vector fields and $\rho$ is a scalar field.

If we want to write Maxwell's equations with differential forms, we need to decide what type of forms will represent $E, B, J$ and $\rho$. Following Stern et al we will decide this based on how these fields are used in various equations.

First, we consider Faraday's law \[ \oint_C E \cdot d\ell = -\frac d {dt} \int_S B \cdot dA \]

$E$ is integrated over a curve, so $E$ naturally corresponds to a 1-form (Since one-forms are objects that you can integrate along curves). We will write the 1-form associated to $E$ as $\eta = E^\flat$. Meanwhile $B$ is integrated over a surface, so $B$ naturally corresponds to a 2-form. We will write this 2-form as $\beta = \star B^\flat$ as the 2-form associated to $B$. $\beta$ can be thought of as a function that measures the flux of $B$ through oriented parallelograms. $J$, the current density, is integrated over surfaces to find the current passing through the surface, so $J$ is naturally a 2-form. We will write this forms as $\mathscr J = \star J^\flat$. Finally, $\rho$ is integrated over volumes to find the enclosed charge, so $\rho$ is naturally a 3-form, and we will call the 3-form $\rho$.

We recall the following rules for translating from vector calculus to differential forms: \[\begin{aligned} (\nabla \cdot v)^\flat &= \star d \star v^\flat\\ (\nabla \times v)^\flat &= \star d v^\flat \end{aligned}\] We can use these rules to write Maxwell's equations in terms of $\eta, \beta, \mathscr J,$ and $\rho$.

We will start with the first equation. On the left hand side, $\nabla \cdot E$ becomes $\star d \star \eta$, which is a 0-form. We want to set it equal to the 3-form $4\pi \rho$. So we use the Hodge star to turn $\star d \star \eta$ into a 3 form and find that $\star \star d \star \eta = 4\pi \rho$. In 3D, $\star\star = 1$, so our equation is just

1'.  $d\star \eta = 4\pi \rho$

Now, we move on to the second equation. $\nabla \cdot B$ becomes $\star d \star (\star \beta)$. Since $\star\star = 1$, this just becomes $\star d \beta$. So our equation is $\star d \beta = 0$. Applying $\star$ to both sides gives $d\beta = 0$.

2'.  $d\beta = 0$

For the third equation, our substitutions gives us $\star d \eta = -\frac 1 c \frac{\partial\star \beta} {\partial t}$. Applying $\star$ to both sides yields $d\eta = -\frac 1 c \star \frac{\partial\star \beta} {\partial t}$. But we can pull the $\star$ inside the derivative to get

3'.  $d \eta = -\frac 1 c \frac{\partial \beta}{\partial t}$

Finally, we translate the last equation. Our substitution rules give us

4'.  $\star d \star \beta = \frac {4\pi} c \star \mathscr J + \frac 1 c \frac{\partial \eta}{\partial t}$


Putting all of the equations together, we have

1'.	$d \star \eta = 4 \pi \rho$
2'.	$d \beta = 0$
3'.	$d \eta = - \frac 1 c \frac{\partial \beta}{\partial t}$
4'.	$\star d \star \beta = \frac{4\pi}{c} \star \mathscr J + \frac 1 c \frac {\partial \eta}{\partial t}$

Now, we have written the equations using differential forms. But the equations still don't look very relativistic yet - we still have a big distinction between space and time derivatives. For our next step, we will stop thinking about forms on space that change over time, and instead think about forms on $3+1$-dimensional spacetime (i.e. spacetime with 3 spatial dimensions and 1 time dimension). In spacetime, we have both the spatial exterior derivative, the spatial Hodge star, the spacetime exterior derivative, and the spacetime Hodge star. We will denote the spatial operators $d_s, \star_s$ respectively, and we will denote the spacetime operators $d$ and $\star$.

The Exterior Derivative and Hodge Star in Spacetime

Before we can write Maxwell's equations in spacetime, we have to learn about how our spatial operators are related to our spacetime operators. We will use the convention that coordinates in spacetime are written like \[ (x^0, x^1, x^2, x^3) = (ct, x, y, z) \]

The Exterior Derivative

Now, we will look at the relationship between $d$ and $d_s$. Let $\omega = \sum_I \omega_I dx^I$ be a spatial differential form (i.e. no component of $\omega$ involves a $dx^0$). We can compute $d\omega$ as follows \[\begin{aligned} d\omega &= \sum_I d \omega_i \wedge dx^I\\ &= (dx^0 \wedge \partial_0 + d_s)\omega \end{aligned}\]

So we see that $d = dx^0 \wedge \partial_0 + d_s$. The spacetime exterior derivative is just the spatial exterior derivative with an extra term related to the time derivative.

The Hodge Star of Spatial Forms

Now, we will relate $\star$ and $\star_s$. Let $\omega$ be a spatial $k$-form. The spacetime Hodge star is defined by the property that \[ \omega \wedge \star \omega = \left\langle \omega, \omega\right\rangle \mu \] Here $\mu = dx^0 \wedge dx^1 \wedge dx^2 \wedge dx^3$ is the spacetime volume form. Let $\mu_s = dx^1 \wedge dx^2 \wedge dx^3$ be the spatial volume form. Clearly $\mu = dx^0 \wedge \mu_s$. Furthermore, we know that $\omega \wedge \star_s \omega = \left\langle \omega, \omega\right\rangle \mu_s$. Therefore, \[\begin{aligned} \omega \wedge \star \omega &= \langle \omega, \omega \rangle \mu\\ &= dx^0 \wedge \langle \omega, \omega \rangle \mu_s\\ &= dx^0 \wedge \omega \wedge \star_s\omega\\ &= \omega \wedge (-1)^k (dx^0 \wedge \star_s \omega) \end{aligned}\] So $\star \omega = (-1)^{k} \; dx^0 \wedge \star_s \omega$ when $\omega$ is purely a spatial $k$-form.

The Hodge Star of Forms with a Time Component

Now, suppose that $\omega = dx^0 \wedge \omega_s$ where $\omega_s$ is a spatial form. Then $\left\langle \omega, \omega\right\rangle = -\left\langle \omega_s,\omega_s\right\rangle$, so we need $\omega \wedge \star \omega = -\left\langle\omega_s,\omega_s\right\rangle\mu$. We know that $\omega_s \wedge \star_s \omega_s = \left\langle\omega_s,\omega_s\right\rangle \mu_s$. Thus, \[\omega \wedge \star_s \omega_s = dx^0 \wedge \omega_s \wedge \star_s \omega_s = dx^0 \wedge \mu_s = \mu \] So, $\star \omega = -\star_s \omega$ when $\omega$ is the wedge product of $dx^0$ and a spatial form.

Finally, we note for completeness that $\star dx^0 = -\mu_s$.

Covariant formulation of Maxwell's equations

Finally, we've developed all of the tools we need to write Maxwell's equations in spacetime. We will begin with the homogeneous equations (equations two and three). Maxwell's second equation tells us that $d\beta = 0$ and Maxwell's third equation tells us that $d_s \eta + \frac 1 c \frac{\partial \beta}{\partial t} = 0$. Because, we write coordinates in spacetime as \[ (x^0, x^1, x^2, x^3) = (ct, x, y, z) \] it turns out that $\frac 1 c \frac{\partial \beta}{\partial t} = \frac{\partial \beta}{\partial x^0} =: \partial_0 \beta$. So we can write Maxwell's third equation as $d_s \eta + \partial_0 \beta = 0$

The second equation is an equation of 3-forms, and the third equation is an equation of 2-forms. We can make them both equations of 3-forms by wedging the third equation with $dx^0$. Then we have $d_s \beta = 0$ and $dx^0 \wedge(d_s \eta + \partial_0\beta) = 0$. We can add these together to get $d_s \beta + dx^0 \wedge \partial_0 \beta + dx^0 \wedge d_s\eta = 0$. We note that because we are adding together forms of different types, their sum is 0 if and only if the individual terms in the sum are 0. So this equation expresses both Maxwell's second law and his third law. Inspecting the sum, we see that the first two terms are our expression for $d\beta$! Furthermore, the last term is $d(\eta \wedge dx^0)$. \[\begin{aligned} d(\eta\wedge dx^0) &= d\eta \wedge dx^0\\ &= (d_s \eta + dx^0 \wedge \partial_0 c\eta) \wedge dx^0 \\ &= d_s \eta \wedge dx^0\\ &= dx^0 \wedge d_s \eta \end{aligned}\] Therefore, we can write Maxwell's second and third equations as $d\beta + d(\eta \wedge dx^0) = 0$, or $d(\beta + \eta \wedge dx^0) = 0$. To simplify even more, we call $F := \beta + \eta \wedge dx^0$ the Faraday tensor, and simply write $dF = 0$.

Now, we consider $d \star F$. We'll start by computing $\star F$. \[ \star F = \star(\beta + \eta \wedge dx^0) = \star \beta + \star(\eta \wedge dx^0) \] Since $\beta$ is a spatial 2-form, $\star \beta = dx^0 \wedge \star_s \beta$. Since $\eta$ is a spatial 1-form, \[ \star(\eta \wedge dx^0) = -\star(dx^0 \wedge \eta) = -(- \star_s \eta) = \star_s \eta\] Putting this together shows us that $\star F = dx^0 \wedge \star_s \beta + \star_s \eta$. Now, we can take the exterior derivative. \[\begin{aligned} d \star F &= d(dx^0 \wedge \star_s \beta + \star_s \eta)\\ &= d_s \star_s \eta + dx^0 \wedge (\partial_0 \star_s \eta - d_s \star_s d_s \beta) \end{aligned}\]

Maxwell's first equation tells us that $d_s \star_s \eta = 4\pi \rho$. Maxwell's fourth equation tells us that $\partial_0\eta - \star_s d_s \star_s \beta = -\frac{4\pi} c \star_s \mathscr J$. Therefore, \[\begin{aligned} d\star F &= 4\pi \star \rho\;dx^0 -\frac {4\pi} c dx^0 \wedge \mathscr J\\ \end{aligned}\] We define $c\rho - dx^0 \wedge \mathscr J = \mathfrak J$ to be the four-current. Now, our equation reads $d \star F = \frac {4\pi} c \mathfrak J$. This lets us finally express Maxwell's equations (in cgs units) as \[\begin{aligned} dF = 0 \quad\text{and}\quad d \star F = \frac{4\pi} c \mathfrak J \end{aligned}\]

Final Thoughts

I think this form of Maxwell's equations is very pretty. Because $F$ and $\mathfrak J$ are coordinate-independent objects, you can tell these equations must also be Lorentz covariant. And the equations don't distinguish between space and time coordinates.

Noether's theorem tells us that conserved quantities come from symmetries of physical systems. For example, momentum is conserved because the laws of physics are translation invariant.
This deep insight is helpful for understanding when quantities should be conserved. A mass falling off of a building is allowed to gain momentum because the system is not translation invariant - as you move vertically, the gravitational potential changes. However, a train moving along its tracks should conserve momentum because no relevant physical quantity changes as you move around the surface of the earth.

Proving Noether's Theorem

In the system of Hamiltonian Mechanics, the proof of Noether's theorem is surprisingly simple and elegant. First, we need to set up some machinery. Recall Hamilton's equations of motion \[\begin{aligned} \dot p_i &= \frac{\partial H} {\partial q_i}\\ -\dot q_i &= \frac{\partial H} {\partial p_i} \end{aligned}\] Hamilton's equations define a vector field $X_H = (\dot q, \dot p)$ on phase space that describes how a particle evolves over time. The trajectory of a particle starting at position $q$ with momentum $p$ is the integral curve of $X_H$ passing through point $(q,p)$.
We can express Hamilton's equations more simply using a symplectic form. A symplectic form is a closed, nondegenerate differential 2-form. Using $\Omega$, Hamilton's equations become \[ dH = \iota_{X_H} \Omega \] Where $d$ is the exterior derivative and $\iota_{X_H} \Omega$ is the interior product, a one-form defined by $(\iota_{X_H} \Omega)(X_1) = \Omega(X_H, X_1)$.
Now, let $X_G$ be an infinitesimal symmetry transformation. Then $\mathcal{L}_{X_G}H = 0$. That is to say, if we move space a small amount in the $X_G$ direction, the Hamiltonian stays the same. This is exactly what we mean by a symmetry. Furthermore, let $X_G$ be the symplectic gradient of some potential function $U_G$. i.e. $dU_G = \iota_{X_G} \Omega$. Then \[\begin{aligned} 0 &= \mathcal{L}_{X_G} H\\ &= \iota_{X_G} dH + d \iota_{X_G} H &&\text{Cartan's magic formula}\\ &= \iota_{X_G} dH + 0 &&H\;\text{doesn't take arguments, so}\; \iota_{X_G}H = 0\\ &= \iota_{X_G} \iota_{X_H} \Omega &&\text{definition of}\;X_H \\ &= \Omega(X_H, X_G) && \text{definition of the}\; \iota \; \text{operation}\\ &= -\Omega(X_G, X_H) && \Omega\;\text{is antisymmetric}\\ &= -\iota_{X_H} \iota_{X_G} \Omega && \text{definition of the}\;\iota\;\text{operation}\\ &= -\iota_{X_H} dU_G && \text{definition of}\; X_G\\ &= -\iota_{X_H} dU_G + d\iota_{X_H} U_G &&U_G\;\text{doesn't take arguments, so}\; \iota_{X_H}U_G = 0\\ &= -\mathcal{L}_{X_H} U_G && \text{Cartan's magic formula}\\ \end{aligned}\] Therefore, the quantity $U_G$ does not change when we flow along the vector field $X_H$. But flow along $X_H$ is time evolution! So $U_G$ is a conserved quantity over time!

Examples

Translation in One Dimension

Suppose we have a one-dimensional physical system. Furthermore, suppose our Hamiltonian is invariant under translation. The vector field that infinitesimally moves things in the $x$ direction is the vector field that points in the $x$ direction. We need to express this vector field as a symplectic gradient. So we want a function $U(x, p)$ that satisfies \[\begin{aligned} \frac{\partial U(x, p)} {\partial x} &= 0\\ \frac{\partial U(x, p)} {\partial p} &= -1 \end{aligned}\] Clearly, $U(x, p) = -p$. Satisfies this condition. So the quantity $-p$ (and therefore $p$ as well), is conserved in this physical system! Just like that, we have shown the conservation of momentum!

Rotation in Two Dimensions

Suppose we have a two-dimensional physical system that is invariant under rotation. A rotation by an angle $\theta$ is given by the matrix \[\begin{pmatrix} \cos \theta & -\sin \theta\\\sin \theta & \cos \theta\end{pmatrix}\] For very small $\theta$, this matrix is approximately \[\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}\] Rotation affects both position and momentum the same way. So an infinitesimal rotation is given by the transformation $\dot x = -y, \dot y = x, \dot p_x = -p_y, \dot p_y = p_x$. Now, to express this vector field as a symplectic gradient, we need a function $U(x, y, p_x, p_y)$ satisfying \[\begin{aligned} \frac {\partial U} {\partial x} &= \dot p_x = -p_y & \frac {\partial U} {\partial y} &= \dot p_y = p_x\\ \frac {\partial U} {\partial p_x} &= -\dot x = y & \frac {\partial U} {\partial p_y} &= -\dot y = -x \end{aligned}\] To satisfy these conditions, we pick $U(x, y, p_x, p_y) = yp_x - xp_y$, which is the angular momentum!

In quantum mechanics, we represent a particle as a vector in a 'state space' $V$. If we have two particles, we represent the pair as a vector in a product vector space $V_1 \otimes V_2$. This product space is called the 'tensor product'. But what is this tensor product? And why is does it represent pairs of particles?

Warm Up: The Direct Product

Before we talk about the tensor product of vector spaces, we'll go over a more intuitive way of taking the product of vector spaces. Given vector spaces $V$ and $W$, the direct product, $V \times W$, is defined as the set of all pairs of vectors $(v, w)$ for $v \in V$ and $w \in W$. We add together vectors in this new space component by component. \[ (v_1, w_1) + (v_2, w_2) = (v_1 + v_2, w_1 + w_2) \] And we scale up vectors by scaling up both components \[ \lambda(v, w) = (\lambda v, \lambda w)\]

Tensor Products of Vector Spaces

To get the tensor product $V \otimes W$, we can modify the direct product. We still want to look at pairs $(v, w)$. But we'll change the definitions of multiplication and addition a bit. In our new definition of scalar multiplication, multiplying our vector by a scalar only scales one of the components. \[ \lambda(v, w) = (\lambda v, w) = (v, \lambda w)\] For addition, we now only define addition if one of the components matches. \[ (v_1, w) + (v_2, w) = (v_1 + v_2, w)\] The sum only works because the second component in each term is $w$. We get a similar sum if the first components are equal and the second components are different. For all other sums, we just define them as themselves. $(v_1, w_1) + (v_2, w_2)$ is just defined to be itself. It cannot be simplified.

Finally, instead of writing $(v, w)$, we instead write $v \otimes w$. This way, it looks different from the elements of $V \times W$. We call this new space the tensor product of $V$ and $W$.

Simple Example

Let's look at the tensor product of $\mathbb{R}$ with $\mathbb{R}$. The simplest elements of $\mathbb{R} \otimes \mathbb{R}$ look like $1 \otimes 2 + 3 \otimes 4$. Using our rules we defined earlier, we can do things like

\[\begin{aligned} 2 \otimes 3 + 4 \otimes 6 &= 2 \otimes 3 + 4 \otimes 2\cdot 3\\ &= 2 \otimes 3 + 8 \otimes 3\\ &= 10 \otimes 3 \end{align*}

But Why?

The definition of a tensor product looks fairly arbitrary. But it winds up having some nice properties that turn out to be interesting, and surprisingly natural, to study. In order to describe these properties and why they're useful, we will have to make an expedition into the wonderful land of algebra.

Over the last 100 years, mathematicians have realized that when studying mathematical objects, it is incredibly useful to study functions between these objects. When you're looking at vector space, the natural functions to study are linear functions. A linear function is a function that commutes with addition and scalar multiplication. That is to say, it is a function $f:V \to W$ such that $f(v_1 + v_2) = f(v_1) + f(v_2)$, and $f(\lambda v) = \lambda f(v)$.

A lot of functions that take in multiple arguments also have a similar property. For example, let's look at multiplication of real numbers. Mathematically, we can write this as a function that takes in two numbers and spits one number back out. i.e., $m : \mathbb{R} \times \mathbb{R} \to \mathbb{R}$. If we fix the first number and vary the second number, this is a linear function!

\[\begin{aligned} m(a, b_1 + b_2) &= a(b_1 + b_2)\\ &= ab_1 + ab_2\\ &= m(a, b_1) + m(a, b_2)\\ m(a, \lambda b) &= a\cdot \lambda b\\ &= \lambda(a b)\\ &= \lambda \cdot m(a, b) \end{aligned}\]

But it actually has a stronger property. If we fix the second argument and vary the first argument, we also get a linear function out. So this function $m$ is a linear function in either argument. We call such a function bilinear (or multilinear if it takes more than 2 arguments). Multilinear functions pop up all over the place. The cross product, dot product and determinant are all multilinear.

From the definition of multilinearity, we get some identities that multilinear functions have to satisfy. Suppose $f$ is multilinear. Then

\[\begin{aligned} f(a, \lambda b) &= \lambda f(a, b) = f(\lambda a, b)\\ f(a, b_1 + b_2) &= f(a, b_1) + f(a, b_2)\\ f(a_1 + a_2, b) &= f(a_1, b) + f(a_2, b) \end{aligned}\] Do these equations look familiar? They're exactly the rules that we made the tensor product follow.

Tensor Products: An Algebraic Perspective

To an algebraist, tensor products are fundamentally about linear maps. The tensor product of two vector spaces $V$ and $W$ is defined by a universal property. Suppose $h$ is a function that takes in an element of $V$ and an element of $W$ and returns an element of a third vector space $Z$. We can write this as a function $h:V \times W \to Z$. Furthermore, let $h$ be bilinear. Then we can extend $h$ to a unique linear map $\bar h$ from $V \otimes W$ to $Z$. If we want to be fancy, we can draw this requirement as the following commutative diagram

The map $\varphi:V \times W \to V \otimes W$ is in a sense, the 'most general' bilinear map out of $V \times W$. We can write any bilinear map $V \times W \to Z$ as the composition of $\varphi$ with some linear map $V \otimes W \to Z$.

This is what tensor products are really about. The confusing definition given above is made specifically so that tensor products play nicely with bilinear maps. Whenever you see a tensor product, you should look around for bilinear maps.

Back to Quantum Mechanics

In quantum mechanics, linear functions play a central role. You look at quantum states as vectors in some large vector space, and special linear functions on this vector space correspond to quantities you can observe. These are called "observables".

Now, what if we have two particles? Each one independently is a vector in some vector space, but how do we describe the pair? Suppose we have some observable that we can measure on the pair. When restricted to one particle, it should be a linear operator like an ordinary observable. So really, our observables for the pair should be bilinear functions from the direct product of the vector spaces. This means that they are linear operators on the tensor product space!

The commutative diagram is from wikipedia

1 What is a Category?

A category is just a bunch of dots with arrows going between them. The tricky part is interpreting what those dots and arrows mean.

In general, there are three conditions that these dots and arrows need to satisfy. Every dot needs a special arrow that goes from the dot to itself. We call this special arrow the identity.
Also, if we have an arrow from dot $A$ to dot $B$ and an arrow from dot $B$ to dot $C$, then we have to be able to combine these arrows to get an arrow from dot $A$ to dot $C$.
Finally, combining an arrow with the identity arrow shouldn't change it.

And that's all a category is. Just a collection of dots and arrows satisfying these three rules.

2 What Things Are Secretly Categories?

One simple category is Set, the category of sets. The dots in Set are, of course, sets. The arrows are functions between sets. Every set has an identity map to itself, functions between sets can be composed, and composing a function with the identity map does nothing. So the arrows in Set follow our rules.

A lot of categories follow this model, where the dots are some sort of collection and the arrows are functions between them. For example, the dots in the category Grp are groups and the arrows are group homomorphisms. Every group has an identity homomorphism to itself, the composition of two homomorphisms is a homomorphism, and composing a homomorphism with the identity does nothing. So Grp is also a category.

Along the same lines, we have Ring, the category of rings and homomorphism between them, and Top, the category of topological spaces and continuous maps between them. The list goes on.

But not every category is of this form. As a more exotic example, you can look at a group as a one-dot category. Each element of the group corresponds to an arrow in the category. The identity element of the group is the identity arrow. And combining two arrows corresponds to the group operation. This object is still a category, but in this case it doesn't make sense to view the dot as a set and the arrows as structure-preserving transformations.

3 Why Is This A Useful Notion?

To give some motivation, let's just consider Grp for the moment. The categorical viewpoint is that instead of studying how the individual elements of a group fit together, we should study structure-preserving transformations between groups. Often, when we look at a group, we want to know about its subgroups or quotient groups. But since every subgroup is the image of a homomorphism into the group, we can just study subgroups by studying homomorphisms into the group. And by the first isomorphism theorem, every quotient of a group is the image of a homomorphism out of the group. So we can really study subgroups and quotient groups just by studying group homomorphisms.

The idea of studying structured objects by studying maps that preserve that structure pops up all over in mathematics. Category theory just takes this idea to the extreme by studying only these maps and forgetting everything else about the objects.

The first image is from here

We see ice melting and water boiling every day. But why does it happen? If you think about it, it's kind of weird that the properties of water can change so suddenly. Today, I'm going to talk about why this happens. But first, we have to come up with some equations that we can use to describe the properties of gases.

The Van der Waals Equation

At some point in chemistry class, you might have seen the ideal gas law, $PV = nRT$. An ideal gas is a theoretical gas whose particles take up no volume and don't interact with each other. These properties make ideal gases easy to do math about, and give us nice equations like the ideal gas law. In fact, if we change our units, we can get an even nicer ideal gas law. If we let $N$ be the number of particles and $\tau$ be the temperature in joules, then instead of the normal chemistry ideal gas law, we just have $PV = N\tau$. It's nice and simple.

Unfortunately, if we want to study phase transitions, this ideal gas model is a little bit too simple. Many phase transitions happen because of interactions between molecules, which cannot happen in an ideal gas. If we want to understand why water boils, we'll need a more sophisticated model. We can add in some correction terms to make two of our assumptions a bit more realistic. First of all, instead of assuming our particles are point masses, we can instead give them each a little volume. This decreases the amount of empty space in the container of gas. Furthermore, how much the empty space decreases should depend linearly on the number of gas particles. So we can replace $V$ in our equation by $V - Nb$ for some positive constant $b$. Next, we add in an attractive force between the particles. For particles in the middle of the gas, this doesn't do much. They have particles surrounding them on all sides, so the attractive forces in all directions cancel each other out. But it does affect the particles near the edges of the container. They are pulled back towards the middle of the container since almost all of the other particles are in that direction. The magnitude of this force depends on the number density ($\frac N V$) of particles in the container. The number of particles at the edge of the container is also proportional to the number density of particles. This means that the pressure of our non-ideal gas is decreased by $a \frac {N^2}{V^2}$ for some positive constant $a$. These two modifications give us the Van der Waals equation \[ \left(p + \frac {N^2}{V^2} a\right) (V - Nb) = N \tau \] The Van der Waals equation is still a pretty crude approximation and still only works for dilute gases, but it will allow us to understand phase transitions qualitatively.

Phase Transitions

So, now we've got a fancy new equation to model non-ideal gases. Let's see what it tells us. We'll begin by looking at how pressure varies with volume. I picked some arbitrary $a, b$ and $N$ values and plotted $V$ vs $p$ for various temperatures. The plots look like this.

For $\tau$ below some critical temperature, we see that pressure first dips down, then goes up a bit, and then goes back down again as volume decreases. This is weird. Intuitively, if you squish a substance, the pressure should go up. But there's a region in the plot where decreasing the volume decreases the pressure. As you squish the material more, it resists you less. This seems pretty unrealistic. What's going on there?

To analyze this weird behavior, we need to think about the energy stored by squishing the gas. At some point in physics class, you might have seen that $W = \int F\;dx$. Work (energy) is a force applied over a distance. If we multiply through by $\frac{area}{area}$, we get another form of the equation that is often more useful with gases. \[ W = \int F \cdot \frac{area}{area}\;dx = \int \frac{F}{area} \; d(x \cdot area) = \int p\;dV \] So $\int p \; dV$ can be seen as the energy stored in the system. In fact, if our system is at a constant temperature, then $\int p \; dV$ is the Helmholtz free energy of the system. We denote the Helmholtz free energy by $A$.

Now, we'll try to use the Helmholtz free energy to understand what goes on in the weird region of the graph we identified above. We'll assume that our temperature is constant, so we have that the Helmholtz free energy when the system has volume $v$ is $A(v) = -\int^\infty_v p\;dV$ That weird dip in our $p-V$ graph makes the Helmholtz free energy flatten out a little bit in that area. This means that in that region, the Helmholtz free energy is not a convex function of $V$. In the following plot, I exaggerate the effect a bit, but it makes it a lot easier to see what happens next.

Systems don't like to have high energy, and gases are no exception. A gas tries to minimize its Helmholtz free energy however it can. This is what makes Helmholtz free energy a useful topic to study. And by using a neat trick, a gas can 'cheat' to get a lower free energy than our plot above would predict by taking advantage of all the weirdness that was confusing us earlier. Our plot describes the Helmholtz free energy of a homogenous substance. But a gas doesn't have to be homogenous. Some of it could be in one state, and some could be in another. Normally, this is not a helpful thing to do, so gases don't do it. But because of our weird graph, a gas can use this trick to lower its free energy.

If part of the gas is in the red state and part of it is in the blue state, then the Helmholtz free energy of the gas is a weighted average of the red energy and the blue energy. This means that by splitting itself into two states, the gas can follow the black line on the $A-V$ graph instead of the purple one, thus lowering its energy! This is only possible because of the unusual concavity of the graph, which in turn was caused by the weird dip in the $p-V$ graph. But what does this mean? When does a gas randomly split into two states? When it condenses into a liquid!

As a gas condenses, the liquid and gas form can coexist. This is precisely the gas becoming inhomogeneous as a means to lower its free energy.

So there you have it. Phase transitions occur because a material can decrease its energy by splitting into an inhomogeneous combination of states rather than smoothly changing its properties. The sudden, mysterious change from a gas to a liquid is a just trick that gases play to take advantage of little bumps in their energy curves!

The image of condensation came from here.

1 What is a Curried Function?

Currying is a way of taking a function of multiple arguments and returning a series of functions that take one argument. For example, consider the simple two-argument function

 function sum(x, y) {
    return x + y;
  }

If we write sum(1, 2), it gives us 3.

We could instead write it as

 function curried_sum(x) {
    return function (y) {
      return x + y;
    }
  }

Now, we have to call the function differently. Instead of writing sum(1, 2), you write curried_sum(1)(2), but it still gives you 3 as a result.

The obvious question to ask now is "Why on earth would I want to write functions that way?" It seems to add verbosity while still giving you the same end result. In fact, currying functions does give you a very useful tool - it allows you to partially apply your functions.

2 Partially Applied Functions

A partially applied function is what happens when you plug in some arguments to a function, but leave other arguments free. Using the example from above, we have that curried_sum(1) is the function that adds 1 to a number. It's what happens when you take the sum function, but fix 1 as the first argument.

This turns out to be a very useful way of defining new functions. The ++ increment operator present in many languages is essentially just curried_sum(1). But this idea of partially applying functions shows up in many other contexts.

2.1 Indefinite Integrals

In fact, we can actually look at the indefinite integral as a partially applied function. A definite integral is a function that takes two arguments: it takes a function to integrate, and an interval to integrate over. We usually write $\int_a^b f(x)\;dx$, but we could just as easily write definite_integral(f, [a, b]). Now, what happens if we partially apply definite_integral to just the first argument? definite_integral(f) should give us a function that takes in an interval and spits out the integral of f over that interval. And that's pretty much what the indefinite integral does! If $F(x) = \int f(x) \; dx$, then $\int_a^b f(x)\;dx = F(b) - F(a)$.

2.2 Dot Products and Dual Vectors

Now, we'll look at an example from linear algebra. Consider the real vector space $\mathbb{R}^3$. The dual vector space $\left(\mathbb{R}^3\right)^\star$ is the set of linear functions $f : \mathbb{R}^3 \to \mathbb{R}$. The Euclidean dot product on $\mathbb{R}^3$ gives us a bilinear function $g:\mathbb{R}^3 \times \mathbb{R}^3 \to \mathbb{R}^3$. For vectors $u = (u_1, u_2, u_3)$ and $v = (v_1, v_2, v_3)$, we define \[g(u, v) = u \cdot v = \sum_{i = 1}^3 u_i v_i \] Now, suppose we partially apply the dot product. $g(u, -)$ gives us a linear map $\mathbb{R}^3 \to \mathbb{R}$. This is an element of the dual space. So partial application of the dot product allows us to convert vectors to vectors in the dual space! This is a useful tool in tensor analysis (e.g. General Relativity).

2.3 Cross Products

As one last example, we'll look at the cross product of vectors in $\mathbb{R}^3$. Before we talk about the cross product, though, we have to talk about determinants. The determinant of a $3 \times 3$ matrix is signed the volume of the parallelepiped defined by its columns. We can also think of the determinant as taking in 3 vectors and returning the corresponding volume.

If you've seen cross products before, you might have been taught to take the cross product of 2 vectors by taking a determinant. \[ (a_1, a_2, a_3) \times (b_1, b_2, b_3) = \begin{vmatrix} \hat \imath & \hat \jmath & \hat k \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix} \]

Now, how can we make this by partially applying functions? This looks similar to a partially applied determinant. $\det(-, A, B)$ is a function that takes in a vector and returns a scalar. So it lives in the dual vector space. But, similar to what we did above, we can convert the dual vector into a regular vector. And if we do that we get the standard formula for the cross product!

We can also look at this from a different geometric perspective. The triple product of three vectors is the quantity $A \cdot (B \times C)$, and is also equal to the signed volume of the parallelepiped that they define. If we write out the formula for the cross product, we find that \[ (a_1, a_2, a_3) \cdot \begin{vmatrix} \hat \imath & \hat \jmath & \hat k \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{vmatrix} = \begin{vmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{vmatrix}\] So we apply our partially applied function by taking a dot product.

You might have heard the cross product of $A$ and $B$ defined as the vector whose magnitude is the area of the parallelogram defined by $A$ and $B$ and which points perpendicular to both $A$ and $B$. We can understand this geometric definition using what we just learned. The cross product is a partially applied volume function, and you finish applying the function by dotting it with the third vector. The volume of a parallelepiped is the volume of the base times the height. If the magnitude of the cross product is the area of the base, and it points in the direction of the height, then dotting that with the third vector gives base $\times$ height. Exactly the volume we wanted!