Network Theory (Part 33)

Last time I came close to describing the ‘black box functor’, which takes an electrical circuit made of resistors

and sends it to its behavior as viewed from outside. From outside, all you can see is the relation between currents and potentials at the ‘terminals’—the little bits of wire that poke out of the black box:

I came close to defining the black box functor, but I didn’t quite make it! This time let’s finish the job.

The categories in question

The black box functor

\blacksquare : \mathrm{ResCirc} \to \mathrm{LinRel}

goes from the category \mathrm{ResCirc}, where morphisms are circuits made of resistors, to the category \mathrm{LinRel}, where morphisms are linear relations. Let me remind you how these categories work, and introduce a bit of new notation.

Here is the category \mathrm{ResCirc}:

• an object is a finite set;

• a morphism from X to Y is an isomorphism class of cospans

in the category of graphs with edges labelled by resistances: numbers in (0,\infty). Here we think of the finite sets X and Y as graphs with no edges. We call X the set of inputs and Y the set of outputs.

• we compose morphisms in \mathrm{ResCirc} by composing isomorphism classes of cospans.

And here is the category \mathrm{LinRel}:

• an object is a finite-dimensional real vector space;

• a morphism from U to V is a linear relation R : U \leadsto V, meaning a linear subspace R \subseteq U \times V;

• we compose a linear relation R \subseteq U \times V and a linear relation S \subseteq V \times W in the usual way we compose relations, getting:

SR = \{(u,w) \in U \times W : \; \exists v \in V \; (u,v) \in R \mathrm{\; and \;} (v,w) \in S \}

In case you’re wondering: I’ve just introduced the wiggly arrow notation

R : U \leadsto V

for a linear relation from U to V, because it suggests that a relation is a bit like a function but more general. Indeed, a function is a special case of a relation, and composing functions is a special case of composing relations.

The black box functor

Now, how do we define the black box functor?

Defining it on objects is easy. An object of \mathrm{ResCirc} is a finite set S, and we define

\blacksquare{S} = \mathbb{R}^S \times \mathbb{R}^S

The idea is that S could be a set of inputs or outputs, and then

(\phi, I) \in \mathbb{R}^S \times \mathbb{R}^S

is a list of numbers: the potentials and currents at those inputs or outputs.

So, the interesting part is defining the black box functor on morphisms!

For this we start with a morphism in \mathrm{ResCirc}:

The labelled graph \Gamma consists of:

• a set N of nodes,

• a set E of edges,

• maps s, t : E \to N sending each edge to its source and target,

• a map r : E \to (0,\infty) sending each edge to its resistance.

The cospan gives maps

i: X \to N, \qquad o: Y \to N

These say how the inputs and outputs are interpreted as nodes in the circuit. We’ll call the nodes that come from inputs or outputs ‘terminals’. So, mathematically,

T = \mathrm{im}(i) \cup \mathrm{im}(o) \subseteq N

is the set of terminals: the union of the images of i and o.

In the simplest case, the maps i and o are one-to-one, with disjoint ranges. Then each terminal either comes from a single input, or a single output, but not both! This is a good picture to keep in mind. But some subtleties arise when we leave this simplest case and consider other cases.

Now, the black box functor is supposed to send our circuit to a linear relation. I’ll call the circuit \Gamma for short, though it’s really the whole cospan

So, our black box functor is supposed to send this circuit to a linear relation

\blacksquare(\Gamma) : \mathbb{R}^X \times \mathbb{R}^X \leadsto \mathbb{R}^Y \times \mathbb{R}^Y

This is a relation between the potentials and currents at the input terminals and the potentials and currents at the output terminals! How is it defined?

I’ll start by outlining how this works.

First, our circuit picks out a subspace

dQ \subseteq \mathbb{R}^T \times \mathbb{R}^T

This is the subspace of allowed potentials and currents on the terminals. I’ll explain this and why it’s called dQ a bit later. Briefly, it comes from the principle of minimum power, described last time.

Then, the map

i: X \to T

gives a linear relation

S(i) : \mathbb{R}^X \times \mathbb{R}^X \leadsto \mathbb{R}^T \times \mathbb{R}^T

This says how the potentials and currents at the inputs are related to those at the terminals. Similarly, the map

o: Y \to T

gives a linear relation

S(o) : \mathbb{R}^Y \times \mathbb{R}^Y \leadsto \mathbb{R}^T \times \mathbb{R}^T

This says how the potentials and currents at the outputs are related to those at the terminals.

Next, we can ‘turn around’ any linear relation

R : \mathbb{R}^Y \times \mathbb{R}^Y \leadsto \mathbb{R}^T \times \mathbb{R}^T

to get a relation

R^\dagger : \mathbb{R}^T \times \mathbb{R}^T \leadsto \mathbb{R}^Y \times \mathbb{R}^Y

defined by

R^\dagger = \{(\phi',-I',\phi,-I) : (\phi, I, \phi', I') \in R \}

Here we are just switching the input and output potentials, but when we switch the currents we also throw in a minus sign. The reason is that we care about the current flowing in to an input, but out of an output.

Finally, one more trick: given a linear subspace

L \subseteq V

of a vector space V we get a linear relation

1|_L : V \leadsto V

called the identity restricted to L, defined like this:

1|_L = \{ (v, v) :\; v \in L \} \subseteq V \times V

If L is all of V this relation is actually the identity function on V. Otherwise it’s a partially defined function that’s defined only on L, and is the identity there. (A partially defined function is an example of a relation.) My notation 1|_L is probably bad, but I don’t know a better one, so bear with me.

Let’s use all these ideas to define

\blacksquare(\Gamma) : \mathbb{R}^X \times \mathbb{R}^X \leadsto \mathbb{R}^Y \times \mathbb{R}^Y

To do this, we compose three linear relations:

1) We start with

S(i) : \mathbb{R}^X \times \mathbb{R}^X \leadsto \mathbb{R}^T \times \mathbb{R}^T

2) We compose this with

1|_{dQ} : \mathbb{R}^T \times \mathbb{R}^T \leadsto \mathbb{R}^T \times \mathbb{R}^T

3) Then we compose this with

S(o)^\dagger : \mathbb{R}^T \times \mathbb{R}^T \leadsto \mathbb{R}^Y \times \mathbb{R}^Y

Note that:

1) says how the potentials and currents at the inputs are related to those at the terminals,

2) picks out which potentials and currents at the terminals are actually allowed, and

3) says how the potentials and currents at the terminals are related to those at the outputs.

So, I hope all makes sense, at least in some rough way. In brief, here’s the formula:

\blacksquare(\Gamma) = S(o)^\dagger \; 1|_{dQ} \; S(i)

Now I just need to fill in some details. First, how do we define S(i) and S(o)? They work exactly the same way, by ‘copying potentials and adding currents’, so I’ll just talk about one. Second, how do we define the subspace dQ? This uses the principle of minimum power.

Duplicating potentials and adding currents

Any function between finite sets

i: X \to T

gives a linear map

i^* : \mathbb{R}^T \to \mathbb{R}^X

Mathematicians call this linear map the pullback along i, and for any \phi \in \mathbb{R}^T it’s defined by

i^*(\phi)(x) = \phi(i(x))

In our application, we think of \phi as a list of potentials at terminals. The function i could map a bunch of inputs to the same terminal, and the above formula says the potential at this terminal gives the potential at all those inputs. So, we are copying potentials.

We also get a linear map going the other way:

i_* : \mathbb{R}^X \to \mathbb{R}^T

Mathematicians call this the pushforward along i, and for any I \in \mathbb{R}^X it’s defined by

\displaystyle{ i_*(I)(t) = \sum_{x \; : \; i(x) = t } I(x) }

In our application, we think of I as a list of currents entering at some inputs. The function i could map a bunch of inputs to the same terminal, and the above formula says the current at this terminal is the sum of the currents at all those inputs. So, we are adding currents.

Putting these together, our map

i : X \to T

gives a linear relation

S(i) : \mathbb{R}^X \times \mathbb{R}^X \leadsto \mathbb{R}^T \times \mathbb{R}^T

where the pair (\phi, I) \in \mathbb{R}^X \times \mathbb{R}^X is related to the pair (\phi', I') \in \mathbb{R}^T \times \mathbb{R}^T iff

\phi = i^*(\phi')

and

I' = i_*(I)

So, here’s the rule of thumb when attaching the points of X to the input terminals of our circuit: copy potentials, but add up currents. More formally:

\begin{array}{ccl} S(i) &=& \{ (\phi, I, \phi', I') : \; \phi = i^*(\phi') , \; I' = i_*(I) \} \\ \\ &\subseteq& \mathbb{R}^X \times \mathbb{R}^X \times \mathbb{R}^T \times \mathbb{R}^T \end{array}

The principle of minimum power

Finally, how does our circuit define a subspace

dQ \subseteq \mathbb{R}^T \times \mathbb{R}^T

of allowed potential-current pairs at the terminals? The trick is to use the ideas we discussed last time. If we know the potential at all nodes of our circuit, say \phi \in \mathbb{R}^N, we know the power used by the circuit:

P(\phi) = \displaystyle{ \sum_{e \in E} \frac{1}{r_e} \big(\phi(s(e)) - \phi(t(e))\big)^2 }

We saw last time that if we fix the potentials at the terminals, the circuit will choose potentials at the other nodes to minimize this power. We can describe the potential at the terminals by

\psi \in \mathbb{R}^T

So, the power for a given potential at the terminals is

Q(\psi) = \displaystyle{ \frac{1}{2} \min_{\phi \in \mathbb{R}^N \; : \; \phi|_T = \psi} \sum_{e \in E} \frac{1}{r_e} \big(\phi(s(e)) - \phi(t(e))\big)^2 }

Actually this is half the power: I stuck in a factor of 1/2 for some reason we’ll soon see. This Q is a quadratic function

Q : \mathbb{R}^T \to \mathbb{R}

so its derivative is linear. And, our work last time showed something interesting: to compute the current J_x flowing into a terminal x \in T, we just differentiate Q with respect to the potential at that terminal:

\displaystyle{ J_x = \frac{\partial Q(\psi)}{\partial \psi_x} }

This is the reason for the 1/2: when we take the derivative of Q, we bring down a 2 from differentiating all those squares, and to make that go away we need a 1/2.

The space of allowed potential-current pairs at the terminals is thus the linear subspace

dQ = \{ (\psi, J) : \; \displaystyle{ J_x = \frac{\partial Q(\psi)}{\partial \psi_x} \} \subseteq \mathbb{R}^T \times \mathbb{R}^T }

And this completes our precise description of the black box functor!

The hard part is this:

Theorem. \blacksquare : \mathrm{ResCirc} \to \mathrm{LinRel} is a functor.

In other words, we have to prove that it preserves composition:

\blacksquare(fg) = \blacksquare(f) \blacksquare(g)

For that, read our paper:

• John Baez and Brendan Fong, A compositional framework for passive linear networks.

This entry was posted on Tuesday, November 4th, 2014 at 2:19 am and is filed under mathematics, networks, physics. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

One Response to Network Theory (Part 33)

  1. John Baez says:
    10 November, 2014 at 8:58 pm

    Whoops! I screwed up the second step of how to get a linear relation from a circuit made of resistors. I wrote:

    To do this, we compose three linear relations:

    1) We start with

    S(i) : \mathbb{R}^X \times \mathbb{R}^X \leadsto \mathbb{R}^T \times \mathbb{R}^T

    2) We compose this with

    1|_{dQ} : \mathbb{R}^T \times \mathbb{R}^T \leadsto \mathbb{R}^T \times \mathbb{R}^T

    3) Then we compose this with

    S(o)^\dagger : \mathbb{R}^T \times \mathbb{R}^T \leadsto \mathbb{R}^Y \times \mathbb{R}^Y

    In the second step I used the power function

    Q : \mathbb{R}^T \to \mathbb{R}

    to get a linear relation

    1|_{d Q} : \mathbb{R}^T \times \mathbb{R}^T \leadsto \mathbb{R}^T \times \mathbb{R}^T

    But Brendan pointed out that this relation is not the right one. Quite concretely, we have

    1|_{dQ} = \{ (\psi, J, \psi, J) : J_x = \frac{\partial Q}{\partial \psi_x} \} \subseteq \mathbb{R}^T \times \mathbb{R}^T \times \mathbb{R}^T \times \mathbb{R}^T

    What we really want is another relation which I’ll call R_Q. This is

    R_Q = \{ (\psi, I, \psi, I+J) : J_x = \frac{\partial Q}{\partial \psi_x} \} \subseteq \mathbb{R}^T \times \mathbb{R}^T \times \mathbb{R}^T \times \mathbb{R}^T

    In his writeup, Brendan arrives at this in a nice conceptual way. I was trying to take a shortcut, and I went the wrong way!

    Let’s see how this stuff works in an example.

    First let’s do a very simple circuit: a circuit with one input and one output, where the graph has one node and no edges. This is a degenerate case, so it may be sort of confusing, but it’s fundamental. This circuit is an identity morphism in the category \mathrm{ResCirc}. Remember, a morphism in here is a cospan

    In our example now, i and o are identity morphisms, and X = \Gamma = Y is a graph with one node and no edges—which is another way to think about a set with one point. This one-point set is also the set of terminals, T. It takes one number \psi to describe the potential at this point and one number J to describe the current flowing into this point from the input.

    Since i and o are identity morphisms, it’s easy to check that steps 1) and 3) of my three-step procedure give identity morphisms. The only problem is step 2), so let’s look at that.

    Since \Gamma has no edges, no power is consumed by this circuit:

    Q = 0

    So, my recipe gives

    1|_{dQ} = \{ (\psi, 0, \psi, 0) : \subseteq \mathbb{R} \times \mathbb{R} \times \mathbb{R} \times \mathbb{R}

    This is not what we want! It correctly says that the potential at the input equals the potential at the output (since they’re the same point). But it incorrectly says the current at input and output are zero.

    If we work out R_Q, we get

    R_Q = \{ (\psi, I, \psi, I) \} \subseteq \mathbb{R} \times \mathbb{R} \times \mathbb{R} \times \mathbb{R}

    and this is indeed the identity linear relation on \mathbb{R} \times \mathbb{R}.

    Reply

You can use Markdown or HTML in your comments. You can also use LaTeX, like this: $latex E = m c^2 $. The word 'latex' comes right after the first dollar sign, with a space after it.

Fill in your details below or click an icon to log in:

Gravatar
WordPress.com Logo

You are commenting using your WordPress.com account. ( Log Out /  Change )

Google+ photo

You are commenting using your Google+ account. ( Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. ( Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. ( Log Out /  Change )

w
Cancel

Connecting to %s