How do you determine the dimensions of a Lie group? Why is the dimension of $O(n)$ the same as $SO(n)$? Similarly for $\text{GL}(n,F)$ and $\text{GL}^+(n,F)$, where $F$ is a field?

To consider and answer these questions, we first need to define dimension in this context. The **dimension** of an Lie group corresponds to the dimension its accompanying group manifold. Now, the Lie groups that we will consider here will be constructed with elements of $\text{Mat}_n(\mathbb{F})$.

**Definition: **$\text{Mat}_n(\mathbb{F})$ denotes the set of $n\times n $ matrices with entries belonging in $\mathbb{F}=\mathbb{R}$ or $\mathbb{C}.$

$\text{Mat}_n(\mathbb{F})$ cannot be a Lie group since (generally) matrices do not always have an inverse, which is why we consider the **general linear group** defined by

\begin{align} \text{GL}(n,\mathbb{F}) &= \{\mathbb{M}\in\text{Mat}_n(\mathbb{F})\ | \det\mathbb{M} \ne 0\},\\\text{GL}^+(n,\mathbb{F}) &= \{\mathbb{M}\in\text{Mat}_n(\mathbb{F})\ | \det\mathbb{M} > 0\},\\\text{GL}^-(n,\mathbb{F}) &= \{\mathbb{M}\in\text{Mat}_n(\mathbb{F})\ | \det\mathbb{M} < 0\}. \end{align}

The general linear group provides us with two key aspects;

- All elements are invertible matrices. Since matrix structure is well understood, this naturally gives us a formula for the dimension $-$ which is the dimension of the matrix.
- All Lie groups studied from here on out are subgroups of $\text{GL}(n,\mathbb{F}).$

Since we will be using the general linear group to construct all other Lie groups, will list these groups here;

\begin{align}\text{SL}(n,\mathbb{F})&=\{\mathbb{M}\in\text{GL}(n,\mathbb{F}) | \det\mathbb{M} =1 \}\\ \text{O}(n,\mathbb{R})&=\{\mathbb{M}\in\text{GL}(n,\mathbb{R})\ | \mathbb{M}^\top\mathbb{M} = \mathbb{M}\mathbb{M}^\top=\mathbb{I} \}\\ \text{U}(n,\mathbb{C})&=\{\mathbb{M}\in\text{GL}(n,\mathbb{C})\ | \mathbb{M}^\dagger\mathbb{M} = \mathbb{M}\mathbb{M}^\dagger=\mathbb{I} \}\\ \text{SO}(n,\mathbb{R})&= \{\mathbb{M}\in\text{O}(n,\mathbb{R})\ | \det\mathbb{M}=+1 \}\\ \text{SU}(n,\mathbb{C})&= \{\mathbb{M}\in\text{U}(n,\mathbb{C})\ | \det\mathbb{M}=+1 \}\\\end{align}

To visualize how these groups are nested within one another, I’ve created the following diagram:

The idea is simple; create a grid with the $n\times n$ matrices on the horizontal axis and the determinants of those matrices on the vertical axis. The right half plane has the field $\mathbb{F}=\mathbb{R},$ which obviously is a subset of the complex field describing the entire plane. Now we can read off the Lie subgroups immediately. In the right half plane, the top and bottom quadrants are split by the positive and negative determinant of $\text{GL}(n,\mathbb{R})$. This shows us some limitations of the real general linear group in that the manifold that accompanies the Lie group is not simply connected. Instead there are two connected components depending on the determinant of the real general linear group. This constraint however is not applicable to $\text{GL}(n,\mathbb{C})$, as this Lie group is simply connected. Here we also see that $O(n)$ and $U(n)$ are not simply connected but have two connected pieces each.

Thankfully, even with the varied Lie subgroups, the procedure to determine the dimension of any matrix Lie group is the same.

\begin{equation}\text{dim} = \text{Total Equations}\ -\ \text{# of Constraints} \end{equation}

**Note**: Often, I’d prefer to think of this values as not the dimension of the group but instead as the **degrees of freedom of the group**.

So, starting with either of the connected pieces of the real general linear group, we find that there are a total of $n^2$ questions $-$ one for each entry of the $n\times n$ matrix $-$ and there are no constraints in either of those regions. Therefore

\begin{equation}\dim(\text{GL}^+(n,\mathbb{R}))=\dim(\text{GL}^-(n,\mathbb{R}))=n^2.\end{equation}

Likewise, since any complex matrix can be written as a linear combination of its real and imaginary parts, we immediately obtain

\begin{equation}\dim(\text{GL}(n,\mathbb{C}))=2\cdot n^2.\end{equation}

Next, the special linear group in both fields has one constraint, that being the value of the determinant. Following the same procedure as shown above, we have

\begin{align}\dim(\text{SL}(n,\mathbb{R}))&=n^2\ -\ 1,\\ \dim(\text{SL}(n,\mathbb{C}))&=2\cdot n^2\ -\ 2.\end{align}

With $O(n)$ and $U(n)$, we run into some issues. First, let $\mathbb{M}\in O^-(n)$, then the determinant of this matrix could either be $\pm 1.$

**Proof:** $\det(\mathbb{M}^\top\mathbb{M})=\det(\mathbb{I})=1.$ However, since the transpose doesn’t affect determinants, $\det(A^\top) = \det(A)$, we have that $\det(\mathbb{M})^2 = 1\implies$ $ \det(\mathbb{M})=\pm 1.$

So $O(n)$ has two connected pieces just like the real general linear group. However, this time the matrices with negative determinants do not form a group as this section of $O(n)$ is not closed.

**Proof: **Suppose $M,N\in O^-(n)$, that is, $\det(M) = \det(N) = -1$. Then the product $\det(MN) = \det(M)\det(N) = +1$ is not in $O^-(n).$ In fact, this product is an element of $SO(n).$

This is the reason why $SO(n)$ and $O(n)$ have the same dimension. Determining the dimension is a little tricky. Again there are a total of $n^2$ elements. This time, however, the number of constraint equations are as follows;

The diagonal of a matrix in $O(n)$ is given by $(\mathbb{M}\mathbb{M}^\top)_{i=j} = \delta_{i=j} = 1$ and there are $n$ equations here. Since $(\mathbb{M}\mathbb{M}^\top)_{i\ne j}=(\mathbb{M}^\top\mathbb{M})_{i\ne j}=\delta_{i\ne j}=0$, the sum of all diagonals above (or below) the main diagonal contribute to the constraint equations, so $1+2+…+(n-2)+(n-1) = \frac{n(n-1)}{2}$ additional constraints. Therefore the dimension of $O(n)$, and $SO(n)$, is $$\dim(O(n)) = n^2\ -\ n\ -\ \frac{n(n-1)}{2} = \frac{n(n-1)}{2}.$$

Moving onto $U(n)$, a similar issue occurs. First, we note that if $\mathbb{N} \in U(n)$, then $\det(\mathbb{N})=e^{i\theta}.$

**Proof:** $\det(\mathbb{N}^\dagger\mathbb{N})=\det(\mathbb{I})=1.$ However, since the transpose doesn’t affect determinants, $\det((A^*)^\top) = \det(A^*)$, we have that $\det(\mathbb{N}\mathbb{N}^*) = |\det(\mathbb{N})|^2 = 1\implies$ $ \det(\mathbb{N}) = e^{i\theta}.$

Again for some value of $\theta$, a matrix could have a negative determinant but if two matrices with negative determinant are multiplied together under the group multiplication, these matrices end up in $SU(n).$ But as we will see this does not mean that the dimensions of $U(n)$ and $SU(n)$ are the same.

For a matrix in $U(n),$ we know that the total number of equations is $2\cdot n^2$ which comes from $n^2$ real equations and $n^2$ complex equations. The diagonal of any matrix in $U(n)$ is real, and for this condition we obtain $n$ constraints. However, the off diagonal terms can be broken up again into real and complex parts. For each of these off diagonal triangles, we again have $\frac{n(n-1)}{2}$ constrains for a total of $n(n-1)$ constraints. Finally, we have the dimension of $U(n)$: $$\dim(U(n))=2n^2\ -\ n\ -\ n(n\ -1) = n^2.$$

Finally, and by this points we’re pros at calculating these dimensions, we can find the dimension of $SU(n)$ by realizing the additional constraint here is from the determinant of any matrix in $SU(n)$ being strictly equal to $+1$. Therefore $$\dim(SU(n))=n^2\ -\ 1.$$

Special thanks to Karl Winsor for mathematical guidance.