DEV Community: The Mansions of Science

Frobenius Map Generates the Entire Galois Group

The Mansions of Science — Sun, 21 Jun 2026 00:25:15 +0000

Start with a finite field F_q (meaning there are a finite number of elements, actually exactly q elements). Meaning, any element K of F_q, when raised to the q power, becomes itself again. K^q = K. Make sure q is equal to a prime raised to the some n power, so q = p^n.

To repeat, K = K^q when you do arithmetic in this field.

The Frobenius map FM takes in an element and raises it to the q power. Now, no matter what element R of F_q we input to FM, we get output R^q.

That’s useless.

We already know that R = R^q. So the input equals the output!

BUT! If you take the algebraic closure of F_q, say F_qc, then in general a random element Y of F_qc, when plugged in to FM, will produced Y^q such that Y DOES NOT EQUAL Y^q!

In fact, FM is an element of the Galois group G(F_qc/F_q) and repeated applications of FM generates the entire Galois group.

Finite Depth Quantum Circuits and Symmetry Protected Topological Phases

The Mansions of Science — Sat, 20 Jun 2026 21:56:55 +0000

Suppose you start with a bunch of uncorrelated qubits. This is called the “trivial product state”, state P. Now, you have a desired end state, state Q.

There are 4 cases. You can either use a finite depth quantum circuit (FDQC) to change P to Q. Or, you can’t. You can also choose to use matrices (a quantum gate can represent a matrix) that commute with the symmetry group matrices of Q (called “enforcing symmetry”).

Case 1: Using FDQC (constraint) and enforcing symmetry (constraint), you can still change P to Q. We call Q “trivial”.

Case 2: Using FDQC and enforcing symmetry CANNOT change P to Q. BUT, when you stop using FDQC (you start using any depth), you CAN change to P to Q. We say that Q is a “Symmetry Protected Topological Phase”. P needs to undergo a phase transition to get to Q.

Case 3: Using FDQC but do not enforce symmetry, you can change P to Q. We again call Q “trivial”. The “topology wasn’t intrinsic”. It was held together by the symmetry constraint.

Case 4: Only by NOT using FDQC and NOT using symmetry, you can change P to Q. Q is “intrinsically topological.” “Real” topology.

D-Modules = Derivatives + Vector Space (That’s it)

The Mansions of Science — Sat, 20 Jun 2026 21:44:30 +0000

Let's recap the 12 rules for how to build a vector space is:

You have a set (M) of elements, a set of "vectors"
You have a field (K) of elements, called "scalars" (ex: real numbers, complex numbers, etc.)
The addition operation (+) where vectors A, B in M add up to still be inside M: A + B = C still in M
The multiplication operation (*) where c in K and A in M where c * A = L still in M
The addition operation is associative for three vectors A, B, C in M. (A + B) + C = A + (B + C)
The addition operation commutes: A + B = B + A for vectors A and B in M
There is a zero vector, ZV, in M where ZV + A = A for any vector A in M
Every vector H has an additive inverse AH where AH + H = ZV
Multiplication respects an associative-like rule a * (b * A) = (a * b) * A where a and b are in K and A is in M
If U is the identity element of K, then U * A = A for all A in M
Distributive property holds for adding vectors: c * (A + B) = c * A + c * B where c in K and A, B in M
Distributive property holds for adding scalars: (g + h) * A = g * A + h * A for g, h in K and M

Now, you can replace the field K (or "enlarge" it) with a ring Q. This ring is made up of all linear combinations of x^n and d^n/dx^n. (Linear combinations using complex numbers). Note that Q contains all complex numbers, so if K was the complex field, then we are "enlarging" the field K to the ring Q.

Q is called the Weyl algebra, and if you did replace K with Q, then the previous vector space is now a D-module. Also called "a module 'over' the linear differential operators of one variable." The D historically stands for differential operator.

We say that M is "acted on" by Q. We could also have said M is acted on by K but the "action" of numbers (the elements of a number field, K in this case) is not that interesting.

If you replace a vector space's field with a ring, in general you get a module. Some of the 12 rules will be broken, can you tell which ones? For example, rings are not requires to be commutative.

Conformal Transformations, 2-step Proof that Angles do not Change

The Mansions of Science — Thu, 18 Jun 2026 22:43:41 +0000

Step 1: Write down the limit
Step 2: Use the arc-cosine formula

A complex function, f(z), would assign to each complex number z (z in C) a new complex number, f(z). It is a map from C to C. It sort of deforms C. It is like how f(x) = x^2 deforms the real number line R^1 to look like a parabola.

So in the complex plane C, imagine I have a point A. Suppose there are curves C1 and C2 that intersect at A. When they intersect, both the curves have their own direction. Let's call these directions U and V. We say that a transformation by a complex function f(z) is conformal if the angle between U and V remain unchanged. f(z) is called a conformal map.

Backtrack. Let’s look at point A again. At that point, C1 is passing through. Suppose that the “next point” is point B1. C1, at point A, points from A, toward B1. That direction it is pointing in is the direction U. U is a vector defined as position vector B1 minus position vector A.

Now, C1 is just a bunch of points on the complex plane that, together, trace out a line. C2 is also just a bunch of points on the complex plane that, together, trace out a line.

If you apply f(z) to C, what happens to the points that make up C1 and C2. They also transform. So C1 becomes f(C1) and C2 becomes f(C2). f(C1) is the “image” of C1 under the f(z) transformation.

Now, what does f(B1) - f(A) become? B1 - A use to be the vector U. Now it is the vector f(U). The question is, has the angle between f(U) and f(V) changed?

Suppose f is holomorphic, so that the limit definition of a derivative holds. f’(A) = limit as B1 approaches A (f(B1) - f(A))/(B1 - A). Remember from calculus that the derivative at A is the direction from A to an infinitesimal neighbor A + epsilon. Here we have B1 instead of A + epsilon.

As B1 gets closer to A, we can approximate f(B1) by f(A) + f’(A)(B1 - A). So, f(B1) - f(A) is f’(A)(B1 - A), which is also f’(A)*U.

So we have proved that f(U) = f(B1) - f(A) = f’(A)*(B1 - A) = f’(A)*U. The question is, is the angle between f(U) and f(V), that is to say f’(A)*U and f’(A)*V, the same as the angle between U and V.

It suffices to show that multiplication by a complex number preserves angles.

f’(A) is a complex number. U, remember, is a 2D vector. It pointed from a point A in the complex plane (isomorphic to R^2) to another point B. We can say that U = (u1,u2) = u1 + i*u2. A complex number, here f’(A), can be written as r*e^(iT) (T is theta), or r(cosT + i*sinT).

f’(A)U = r(cosT + i*sinT) * (u1,u2) = r(cosT + i*sinT) * (u1 + i*u2) = r(cosT * u1 - sinT * u2 + i * (sinT*u1 + cosT*u2)).

So the new vector f’(A)*U looks like:

r * [cosT * u1 - sinT * u2]
r * [sinT * u1 + cosT * u2]

If you look very closely, you’ll notice that we can rewrite this as a matrix with entries cosT in the top left, -sinT in the top right, sinT in the bottom left and cosT in the bottom right. Let’s call this matrix K. So U becomes f’(A)*U which can be written as r*K*U.

The angle between two vectors U and V is arccos[(U dot V) / (mag(U)*mag(V))]. The angle between f’(A)*U and f’(A)*V is arccos[(f’(A)*U dot f’(A)*V)) / (mag(f’(A)*U)*mag(f’(A)*V))]. With a lot of effort you can prove that the top is r*r*U dot V. Both K’s go away. Similarly, you can prove that the bottom is r*mag(U)*r*mag(V). Dividing leads to r*r canceling on the top and bottom, leaving us with U dot V over mag(U)*mag(V). So we have proven that the angle between U and V and the angle between f(U) and f(V) are the same.

Note that a conformal map f(z) will preserve the angle between any two vectors at any point in the complex plane.

Stabilizer Subgroups Leave Set Elements Unmoved

The Mansions of Science — Thu, 18 Jun 2026 01:01:43 +0000

Given a group G and a set X, if the action of an element g of G leaves an element x of X fixed, we call g a stabilizer of the point x. The set of all stabilizers of a point x form, in fact, a subgroup. This subgroup is called the stabilizer group of x. If every stabilizer group is trivial, trivial meaning contains only the identity element of G, then we say that the action of G on the set X is “free.” There is no an and b combination, where a is an element of the group G and b is an element of the set X that leaves b unmoved, unless a is the identity element of G.

We say that the action of G on X is free. Free action or free group action or acts freely.

If the action of a group G on a set X is free, then, if you take any element of G (not the identity though) and act on any element x of X, then you will get a distinct element. Distinct means that two different elements will not send x to the same element y in X.

For example, if there are 20 elements in a group G and 500 elements in a set X, if you apply all 20 elements of G to an element x of X, you will get back 20 elements where 1 is x (due to the identity) and the other 19 are distinct.

This is actually used in constructing manifolds and, if you have orbits, which we will discuss next, orbifolds. They are also used in constructing principal G-bundles, which is all the rage nowadays.

Coroots, Killing Form and Langlands Dual Group

The Mansions of Science — Thu, 18 Jun 2026 00:56:43 +0000

Given a semisimple Lie algebra T, we can create an adjoint representation.
Then, the Killing form is a bilinear form B that eats two elements of the Lie algebra and outputs a complex number. B(X,Y) = Trace(adj_X, adj_Y) where X and Y in T. adj_A is defined as adj_A = [A,Z] where A and Z is some element of T. adj_A therefore acts on vector Z from T and we can call it an operator. Composing two such operators and then taking the trace is the definition of the Killing form. So adj_A(Z) = [A,Z], adj_B(Z) = [B,Z] and adj_A(Z) composed with adj_B(Z) is [A,[B,Z]]. The composed action (adj_A composed with adj_B) on a test vector Z is [A,[B,Z]]. The trace of this composed action (which doesn’t involve Z) is the Killing form definition.
Remember, a root L of a Lie algebra T and Cartan Subalgebra (CSA) S means that [H,X] = L(H)*X, where X in T, H in CSA and L(H) is a number. You can say that L is a vector with components L(H_i) (not a vector of T).
If L(H) = B(H_L,H) for all H in the CSA, then we say that H_L is the unique CSA element that corresponds to the root L.
Now, if you plug H_L twice into the Killing form (B(H_L,H_L)), you would get a number. A coroot C_L of the root L is defined as 2*H_L/B(H_L,H_L).

Weights, Roots and Coweights of a Lie group

The Mansions of Science — Wed, 17 Jun 2026 08:24:07 +0000

Suppose you have a Lie group. The generators of this group, along with their commutation relations, form a Lie algebra. The maximal commuting subset of this is called a Cartan Sub-Algebra (CSA).
Suppose you have a CSA with 20 generators.
Given some vector space V, each of the 20 generators can be associated with 20 matrices. If you change V, you change the 20 matrices
If a vector A of the vector space V is an eigenvector of every element of the CSA (let’s call them M1, M2, M3 … M19, M20), then M_i acting on A produces 20 eigenvalues. These 20 numbers, together are called a weight, L. So L = L(e1, e2, e3 … e19, e20). L(M_i) = e_i. M_i acting on A = e_i * A. A is called a weight vector.
In theory, if you are given some random vector space D, you won’t know how many weights, how many L’s, exist in advance in this D-representation of the CSA. In practice, you work with vector spaces that you know something about, so it is possible to predict how many weights you will encounter in this vector space’s representation of the CSA.
So in vector space V there could be 5 weights, L1 L2 L3 L4 L5. In another vector space W there could be 8 weights, N1 N2 N3 N4 N5 N6 N7 N8. In another vector space Y there could be 13 weights, P1-P13.
Ok, now, let’s call the original Lie algebra T and the CSA S. T is, in fact, a vector space. If you represent S using the T-vector space, you have what is called the adjoint representation of this Lie algebra. The adjoint action, instead of being matrix multiplication, is the commutator: Adj_H(X) = [H,X]. H is an element of the CSA S. X is a vector in the Lie algebra T, the T-vector space (since T is a vector space).
Now, for all H in S, a non-zero X in T and [H,X] = number(H)*X, then X is an eigenvector of all CSA (S) elements, H. You can organize these numbers by saying that L(H) = number. L is called a root. L eats a CSA element H and outputs a number. If an element H of the CSA satisfies L_j(H) = integer for all roots L, then H is called a coweight.

Unique Factorization Domains

The Mansions of Science — Wed, 17 Jun 2026 02:42:11 +0000

Last time, we talked about the failure of prime factorization. The set of all numbers of the form (a + b*sqrt(-5)), for example, form a ring, Z[sqrt(-5)]. In this ring, the number 6 factors into 2 and 3 as well as (1-sqrt(-5)) and (1+sqrt(-5)). So unique prime factorization fails because these last two numbers cannot be reduced further.

Now a question is, what negative numbers put inside the square results in unique prime factorization failure? Kurt Heegner was the one who first proposed that only 9 numbers result in NO failure. 1,2,3,7,11,19,43,67,163. No failure means that we still have a unique factorization domain. The expanded ring Z[sqrt(-1)] or Z[sqrt(-2)] still has unique prime factorization for all elements in the ring.

The Failure of Prime Factorization

The Mansions of Science — Wed, 17 Jun 2026 02:38:42 +0000

Look at the number 6. If you only consider the positive integers, then the prime factorization is 2 and 3. That’s it. A unique prime factorization.

Now look at all integers. Now the factorization could be 2 and 3 or -2 and -3. Now suppose you look at all numbers of the form (a + b*sqrt(-5)) where a and b are integers. You can prove that these numbers form a closed set. In fact, it forms the Z[sqrt(-5)] ring. Suddenly, (1-sqrt(-5)) and (1+sqrt(-5)) multiply to 6. Both of these number cannot be broken down any further, so they are prime. 6 no longer has a unique prime factorization. If you set b = 0, you’ll notice that 2 and 3 are also in this ring, so the original unique prime factorization is also here.

This is the beginning of prime ideals (due to people like Dedekind). If you have a ring, which has a bunch of numbers, you can look at subsets, usually infinite subsets of the ring. For example, suppose I have subset A and subset B. Now, although 6 does not have a unique prime factorization, suppose that A*B contains all of its unique prime factorizations, then we say that 6 has a unique prime ideal factorization. And now prime factorization has been “saved” or “re-established.”

Definition of the Cartan Subalgebra

The Mansions of Science — Tue, 16 Jun 2026 09:55:22 +0000

Given an 8 dimensional Lie algebra with generators A B C D E F G H. If it is the case that A B C D all commute with each other, and if you add any of E F G H, you end up breaking commutativity, then the collection A B C D is called a Cartan subalgebra and it has rank 4. For a given Lie group, even this one, it can be proven that all Cartan subalgebras have the same rank. For example, another Cartan subalgebra could very well be E F G H, as long as these 4 commute and adding any of A B C D would break commutativity.

You can say this: The Cartan subalgebra picks out the largest set of directions where flows don’t interfere with each other.

Lie Algebra = Generators + Commutation Relations. That’s it.

The Mansions of Science — Tue, 16 Jun 2026 04:52:37 +0000

So imagine I have a Lie group G. By definition, a Lie group has infinite elements. Out of these infinite number of elements, there are 10 independent generators. Meaning, with these 10 elements, I can "get to" every other element of G. The set of these 10 elements, along with their commutation relations, [X,Y] = XY - YX, where X and Y are two of the generators, form the Lie algebra of this Lie group G. We say that the Lie algebra is 10 dimensional. For example, SO(3) is the group of all rotations in 3D space. It has 3 generators (infinitesimal rotations). So the Lie algebra of the Lie group SO(3) is a 3 dimensional algebra.

We also say that the tangent space at the identity element e is 10 dimensional. There are 10 different independent directions you can “go in”.

Bilinear Form = Function of 2 Vectors (That's it)

The Mansions of Science — Tue, 16 Jun 2026 04:48:44 +0000

Bilinear Form Confusion:

So, bilinear forms use to be confusing to me because I didn’t know what either bilinear meant or what forms meant. But, a form is, very often, just a function. Map is much more common among mathematicians but they’re just functions. They take in inputs and give you outputs. The same functions we’ve been using since middle school.

Now, they take in two vectors and give you back a number. That’s it. The most famous example is the dot product, which even high schoolers know. You take in 2 vectors and you output a number. That’s it.

Bilinear forms are a function that take in 2 vectors and give you back a number.

They have the word “linear” in them because if you plug in (a+b,c), it’s the same as plugging in (a,c) + (b,c). Same for (a,b+c).

If you have a symmetric bilinear form it means that (a,b) = (b,a). Skew-symmetric means what you think antisymmetric means (a,b) = -(b,a). So, like cross products.

Next, I’ll be using bilinear forms to define the symplectic group.