Classical Spaces and Groups

September 2, 2024
By Aathreya Kadambi
Expanding on Lectures by Professor Alexander Givental and Fomenko and Fuchs

What are Classical Spaces?

We recall a few examples of classical spaces here, but more will be discussed throughout this post.

There are linear spaces $\R^n$, $\mathbb{C}^n$, and $\mathbb{H}^n$ (real, complex, and quaternionic) $n$-dimensional spaces.

We call $\{x_1^2 + \dots + x_n^2 \le 1\} := D^n$ the disk and $\partial D^n = S^{n-1}$ the sphere. We can call: $\R^1 \subseteq \R^2 \subseteq \dots \bigcup \R^n =: \R^{\infty}$ so $\R^{\infty}$ is the set of infinite sequence with almost all terms being zero. This contains $D^{\infty}$, the boundary of which is $S^{\infty}$.

The $n$-torus is $T^n = \underbrace{S^1 \times ... S^1}_{n\text{ times}}$.

All of these spaces have topologies on them, which are ideas of closeness and which points are close to each other. This much will be assumed to be familiar to the reader.

What are (Compact) Classical Groups?

Here are the classical groups:

• Real Versions:
  • Orthogonal Group: $O_n := \{V \in \R^{n\times n} : V^{\top}V = I\}$,
  • Special Orthogonal Group: $SO_n = \{V : V^{\top}V = I, \det V = 1\}$.
• Complex Versions: (we can also consider $\mathbb{C}^n$ with hermitian inner product)
  • Unitary Group: $U_n = \{U \in \mathbb{C}^{n\times n} : \overline{U}^\top U = I\}$
  • Special Unitary Group: $SU_n = \{U \in \mathbb{C}^{n\times n} : \overline{U}^\top U = I, \det U = 1\}$
• Quaternionic Versions: (we can consider the inner product $\langle q, q'\rangle = \sum q_i^* q_i'$ where $q_i^*$ is quaternionic conjugation which we will soon see)
  • $Sp_n$: (Compact) Symplectic group

How can we understand the quaternionic versions? As an example, we can consider $Sp_1$:

Example (Understanding $Sp_1$). Consider $\mathbb{H} = \{a + bi + cj + dk\}$ with $i^2 = j^2 = k^2 = ijk = -1$.

Remark. If you go to dublin on some bridge you will actually see that equation written on the bridge. On this bridge this guy was trying invent some kind of multiplication in three dimensional space but it failed so he did it in four dimensional space. I’ll have to look into this later lol.

Cone83, CC BY-SA 4.0, via Wikimedia Commons

The quaternion conjugate of $a + bi + cj + dk$ is $a - bi - cj - dk$. It is not hard to check that

$qq^* = q^* q = a^2 + b^2 + c^2 + d^2.$ The result is a four dimensional division algebra, associative algebra. Maybe there is a better point of view on this, where instead of taking a four-tuple of real numbers, we can think of it as a pair of complex numbers: $\mathbb{H} = \{z + wj : z, w \in \mathbb{C}, j^2 = -1, ji = -ij \}$ I’ll check this later. With this rule, we can compute products: $(x+yj)(z+wj) = (xz - y\overline{w}) + (xw + y\overline{z})j$ You can think of this as working with some two by two matrix: $\begin{bmatrix} x & y \end{bmatrix} \begin{bmatrix} z & w \\ -\overline{w} & \overline{z} \end{bmatrix}$ The second row is actually hermitian conjugate to the first one. Something about a length so you need to multiply by a unit complex number. The length of the first vector being one means the second row will also have length 1. The determinant of this matrix is $z\overline{z} + w \overline{w} = qq^* = \|q\|^2$

Now we can consider unit quaternions, which form a subgroup because of closure. The group $Sp_1$ is a group of unit quaternions: $Sp_1 = \{u \in \mathbb{H} \mid \|u \|= 1\}.$ If you forget about quaternions and just look in complex space, this is the same as $SU_2$ by associating with $z + wj$ this matrix. Geometrically, it is the same as a three-sphere, $S^3$.

$x \mapsto uxu^{-1}$, $\mathbb{H} \rightarrow \mathbb{H}$. $SU_2 \rightarrow SO_3$. Why $\det$ has to be one? That was asked by him but I’m not actually sure what he was talking about.

Strat. The reason the length-preserving maps are also inner product presreving also comes down to just SAS criterion being equivalent to SSS criterion! The inner product is like the angle, and the lengths are norms.

Now we want to keep the rule that linear transformations are written by multiplying by quaternionic matrices. We better introduce multiplication by scalars by right multiplication. That would guarantee that it commutes with multiplying by matrices on the left. There are two notions of right and left vector spaces, and quaternions stick as right vector spaces.

We need to familiarize ourselves with quaternionic spaces.

What we will need to do is a bunch of different types of manifolds. We can think of $O_n$ as the space of orthonormal bases in $\R^n$. These are matrices whose columns form an orthonormal basis. Similarly, $U_n$ will be the orthonomral bases in $\mathbb{C}^n$. We can generalize this though, as orthonormal $k$-frames in $\R^n$, or in $\mathbb{C}^n$, or in $\mathbb{H}^n$.

We call this a Stiefel manifold $V(n,k)$, $\mathbb{C}V(n,k)$, and $\mathbb{H}V(n,k)$. You can transform any frame to another one by orthogonal transformation. $V(n,k) = O_n/O_k.$ The stabilizer of the $k$-frames is $O_k$. The stabilizer looks like: $\begin{bmatrix}A & 0 \\ 0 & B \end{bmatrix}$ where $A$ is some block matrix orthonormal preserving the first one, and $B$ is something preserving the second one. Whenever we have some topological space $X$, we can havea quotient space $X/\sim$. Then there is a canonical projection from $X$ to $X/\sim$.

If we require continuity so that preimages of open sets are open, we can create a topology with the most open sets so that this is continuous. We call this the weak topology. The weakest topology is no topology at all. The maximal supply of open sets is the weakest topology. In general,

What is $V(n, 1)$? It is $S^{n-1}$. $\mathbb{C}V(n,1)$ is the ones so that $|z_1|^2 + \dots + |z_n|^2 = 1$, which becomes $S^{2n-1}$ because it is $\sum x_i^2 + \sum y_i^2 = 1$. Finally, $\mathbb{H}V(n,1) = S^{4n-1}$.

Flag manifolds

We will now talk about these. It is called a flag manifold because it can be drawn as a point, line, nd plane together form a sort of flag. A flag manifold is like: $\{\R^n \supseteq \R^{k_s} \supseteq \dots \supseteq \R^{k_2} \supseteq \R^{k_1}\} = F(n;k_1,k_2;...,k_s)$ You can transform any flag to any flag. What is the stabilizer of a particular flag? Well in $GL_n(R)$, the block of upper triangular matrices will preserve these spaces, so it would be $GL_n(\R)/\{\text{upper diagonal matrices\}} = O_n / O_{k_1} \times O_{k_2 - k_1} \times \dots \times O_{n - k_s}.$

We can also do this in the complex or quaternionic cases.

Complete flags: $F_n = \{\R^1 \subseteq \R^2 \subseteq \dots \subseteq \R^{n-1} \subseteq \R^n\} = F(n, 1,2,...,n-1)$ so the actual flag drawwing is $F_3$. We have that $\dim F_3 = 3$ $\dim F_2 = 1$ and so on continuing as triangular numbers.

Grassman manifolds

$G(n,k)$ or $\mathbb{C}G(n,k)$ or $\mathbb{H}G(n,k)$ is \{\text{k$-dim subspaces in an$n-dim space}\}. This is $O_n / O_k \times O_{n-k} \cong G(n,n-k)$ $GL_n(\R)$ has two connected components with positive and negative determinant. The determinant determines the orientation. Choice of sign for each side of the determinants is the choice of orientaiton (I think). Maybe clarify this in OH.

$G(n,1)$ is also known by definition as projective space: $\R P^{n-1}$. It is called the real projective space of dimension $n-1$. $G(n,1) = \R P^{n-1} = S^{n-1}/\pm 1$ $\mathbb{C} P^{n-1} = S^{2n-1}/U_1$ $\mathbb{H} P^{n-1} = S^{4n-1}/Sp_1$

So we have $\R P^1 = S^1$ $\mathbb{C} P^1 = S^2 = \mathbb{C} \cup \infty$ In the quaternionic case it is also $\mathbb{H} \cup \infty$.

Now we say $G_+(n,k)$ is the $k$-dimensional oriented subspaces in an $n$-dimensional ambient space.