Equality of dimension of $\bigwedge^2(\Bbb R^n)$ and $\text{SO}(n)$

Question

The dimension of $\bigwedge^2(\Bbb R^n)$ as a vector space is $\frac{n(n-1)}{2}$, while the dimension of $\text{SO}(n)$ as a manifold is also $\frac{n(n-1)}{2}$. Even though they don't look exactly the same ($\text{SO}(n)$ is not a vector space), I feel like $\bigwedge^2(\Bbb R^n)$ and $SO(n)$ are somehow related to each other. Is it really the case?

Here is my thought. The vectors in $\bigwedge^2(\Bbb R^n)$ somehow behave like the entries of a skew-symmetric matrix: for any entry $a_{ij}$ in a skew-symmetric matrix, we have $a_{ij}=-a_{ji}$, while for vectors in $\bigwedge^2(\Bbb R^n)$, $e_i\wedge e_j=-(e_j\wedge e_i)$. So we can identify vectors in $\bigwedge^2(\Bbb R^n)$ as skew-symmetric matrices of order $n$ by an isomorphism.

The dimensions of the vector space of skew-symmetric real matrices of order $n$ and $\text{SO}(n)$ are both $\frac{n(n-1)}{2}$. I think this is not a coincidence, as the matrix exponential of a skew-symmetric matrix is an orthogonal matrix. However, I have never seen anyone claiming that the equality of dimensions is due to the exponential relation.

So, do $\bigwedge^2(\Bbb R^n)$ and $\text{SO}(n)$ actually have some subtle relations? Or am I just imagining things, and the equality of dimensions is purely a coincidence?

Edit: What I want to know is if we could actually relate the vectors in $\bigwedge^2(\Bbb R^n)$ and matrices in $\text{SO}(n)$ in a canonical way. For instance, whether we could identify a matrix in $\text{SO}(n)$ with an element in $\bigwedge^2(\Bbb R^n)$.

Edit 2: It happens that the question I have in mind may be related to physics. See this post on Physics Stack Exchange. Sorry for not including the original context.

When I consider $\bigwedge^2(\Bbb R^n)$, it is actually the vector space of all angular momenta, and $\text{SO}(n)$ is the group describing rotational symmetry. What I actually have in mind is that

the equality of the dimensions may be due to that angular momentum is the conserved quantity corresponding to rotational symmetry.

For more details, see the post in Physics Stack Exchange.

Answer 1

You are right saying that $\text{SO}(n)$ is a group, not a vector space. But is has an associated vector space - the tangent space at any point, which is also $\frac{n(n-1)}{2}$-dimensional. Furthemore, the tangent space at identity can be enriched with some algebraic structure - anticommutative multiplication, aka Lie bracket $[x,y]$ - that makes it into the Lie algebra $\mathfrak{so}(n)$. This algebra is isomorphic to the Lie algebra of $n\times n$ skew-symmetric matrices, if the commutator is taken as the Lie bracket: $[x,y]=xy-yx$.

The trick is how to relate $\Lambda^2\left(\mathbb{R}^n\right)$ and $\mathfrak{so}(n)$. We can embed $\Lambda^2\left(\mathbb{R}^n\right)$ into $(\mathbb{R}^n)^{\otimes 2}$, say, via $x \wedge y \mapsto x\otimes y - y\otimes x$, and use the scalar product to identify vectors with forms and identify tensors with linear transformations. This way, $\Lambda^2\left(\mathbb{R}^n\right)$ will get mapped to skew-symmetric linear transformations.

Answer 2

We can attach a vector space of dimension $\binom{n}{2}$ to $SO(n)$, namely its Lie algebra $\mathfrak{so}(n)$, which consists of skew-symmetric matrices of order $n$. Since we have dimension equality, i.e., $\dim SO(n)=\dim \mathfrak{so}(n)=\dim \wedge^2(\mathbb{R}^n)=\binom{n}{2}$, as for all connected Lie groups $G$ with Lie algebra $\mathfrak{g}$ holds $\dim G=\dim \mathfrak{g}$, your claim on the equality of dimensions is correct. For more information on Lie group-Lie algebra correspondence see here.

Answer 3

Hint. The tangent space of $SO(n)$ is exactly the space of skew-symmetric matrices. The same holds of $SU(n)$, whose tangent space is the space of Hermitian matrices. Check it!

Answer 4

Given any Lie group $G$ (say a matrix Lie group for simplicity), any differentiable path through the identity will have a velocity vector. The collection of all of these form the tangent space for $G$ at the point $\mathrm{Id}\in G$, called the lie algebra $\mathfrak{g}$ (it has a binary operation coming from conjugation in $G$), which will be a vector space having the same dimension as $G$.

In 3D all rotations are rotations $R$ around an axis $\ell$. However, this description doesn't generalize to higher dimensions. Instead, think of the plane $\Pi$ perpendicular to the axis: then $R$ acts as a 2D rotation when restricted to $\Pi$ and acts as the identity map restricted to $\ell$. This generalizes: in any real inner product space $V$, given any 2D oriented subspace $\Pi$ and angle $\theta$ we can form a plane rotation $R(\Pi,\theta)$ which acts as a 2D rotation by the angle $\theta$ along the orientation when restricted to a function of $\Pi$, and acts as the identity map on $\Pi$'s orthogonal complement.

In fact, every rotation $R\in\mathrm{SO}(V)$ is expressible as a product

$$R=\prod_i R(\Pi_i,\theta_i) $$

of plane rotations $R(\Pi_i,\theta_i)$, where $\{\Pi_i\}$ is a collection of pairwise orthogonal oriented 2D subspaces of $V$, and $\{\theta_i\}$ is a collection of angles. In fact if the angles are distinct the collection of planes in the decomposition is unique. Since the planes are all orthogonal, all of the plane rotations appearing as factors in the product will commute, so there's no reason to specify order.

This is equivalent to saying $R$'s Jordan block decomposition (using some coordinates) is a bunch of $2\times 2$ rotation matrices (plus maybe a $1\times 1$ block with the entry $1$).

If we view $R(\Pi,t)$ as a function of $t$ for some fixed plane $\Pi$, then it is a one-parameter subgroup, i.e. a group homomorphism $R(\Pi,-):\mathbb{R}\to\mathrm{SO}(V)$, and differentiating at $t=0$ gives

$$ \left.\frac{d}{dt}R(\Pi,t)\right|_{t=0}=E(\Pi)$$

where $E(\Pi)$ acts as a right-angle rotation when restricted to $\Pi$ and acts as the zero map on $\Pi$'s orthogonal complement. (To see this, differentiate the definition of $R(\Pi,t)$ "componentwise.") Compare this with the situation for complex numbers: $R(\mathbb{C},t)$ is multiplication by $e^{it}$, whose derivative at $t=0$ is $i$, and multiplication by $i$ is a right angle rotation.

The corresponding representation of "infinitessimal rotations" $E\in\mathfrak{so}(V)$ is

$$ E=\sum_i c_iE(\Pi_i) $$

where again $\{\Pi_i\}$ is are oriented, pairwise orthogonal, and now $c_i$ are scalars.

Note that $R$ being an orthogonal transformation of $V$ means $\langle Ru,Rv\rangle=\langle u,v\rangle$ for all $u,v\in V$. If we view $R$ as a differentiable function of $t$ with $R(0)=I$ and $R'(0)=E\in\mathfrak{so}(V)$, then differentiating this property gives $\langle Eu,v\rangle+\langle u,Ev\rangle=0$. Using $\langle u,Ev\rangle=\langle E^Tu,v\rangle$, we may write this as $\langle (E+E^T)u,v\rangle=0$ for all $u,v$, which implies $E+E^T=0$, i.e. $E$ is a skew-adjoint transformation. (With matrices, this means $R$ is an orthogonal matrix and $E$ is a skew-symmetric matrix.)

Let's switch gears for a moment. Every element of $\bigwedge^2 V$ is a linear combination of wedges $a\wedge b$, and given any particular wedge $a\wedge b$ we may write decompose $b$ into parallel/perpendicular components with respect to $a$ via $b=b_{\|}+b_\perp$, in which case

$$ a\wedge b=a\wedge (b_{\|}+b_\perp)=a\wedge b_{\|}+a\wedge b_\perp=a\wedge b_\perp $$

since $a\wedge b_{\|}=0$ (we may write $b_{\|}=\lambda a$ so this is $a\wedge \lambda a=\lambda(a\wedge a)=0$), and thus without loss of generality all wedges $a\wedge b$ are with $a,b$ orthogonal. Next, we may write $a=r\hat{a}$ and $b=s\hat{b}$, in which case $a\wedge b=(r\hat{a})\wedge(s\hat{b})=(rs)(\hat{a}\wedge\hat{b})$, so every element of $\bigwedge^2 V$ is a linear combination of $a\wedge b$ with $a,b$ orthonormal, (The same procedure can be done to any element of $\bigwedge^r V$ more generally, which is essentially the Gram-Schmidt process.)

Next, if we depict $a,b$ as directed axes in some 2D plane $\mathrm{span}\{a,b\}$, we may wonder what happens to the element $a\wedge b$ if we rotate it within this plane by an angle of $\theta$ say:

$$(\cos(\theta)a+\sin(\theta)b)\wedge(-\sin(\theta)a+\cos(\theta)b). $$

Distributing, using $a\wedge a,b\wedge b=0$ and $b\wedge a=-a\wedge b$ gives

$$ \cos^2(\theta)(a\wedge b)-\sin^2(\theta)(b\wedge a)=a\wedge b.$$

In other words, the element $a\wedge b\in\bigwedge^2 V$ is unaffected by rotating $\{a,b\}$ within their span. Therefore, given any 2D oriented subspace $\Pi$ of $V$, we may pick an ordered orthonormal basis $\{a,b\}$ for it, and then represent $\Pi$ by $a\wedge b$.

(In this way, the oriented Grassmanian $\widetilde{\mathrm{Gr}}(2,V)$, the manifold whose points are 2D oriented subspaces of $V$, may be embedded as a submanifold of $\bigwedge^2V$. This generalizes to $\bigwedge^r V$ too.)

Putting it all together, we now have a way to describe a canonical vector space isomorphism $\bigwedge^2 V\to\mathfrak{so}(V)$ by sending $a\wedge b\mapsto\mathrm{span}\{a,b\}$ (oriented span) whenever $\{a,b\}$ is an ordered orthonormal set and then extending linearly.

Answer 5

Geometric algebra routinely maps $\bigwedge^2 \mathbb{R}^n\to \mathrm{Spin}(n)\to\mathrm{SO}(n)$ via the exponential map (defined via the usual infinite sum within a Clifford algebra) and then via a quotient by $\mathrm{Z}_2$, the latter as the double cover. Bivectors are a useful way of expressing rotations through this map. So the connection is well-understood, and is used all the time. This works for the indefinite orthogonal groups too.

To answer your final "edit" question: yes, the map is canonical. Given an orthonormal basis, the elements of $\bigwedge^2\mathbb{R}^n$ is interpreted in the canonical way as the exterior square of ${R}^n$. The matrix representation of the final rotation of any vector can be determined in terms of the original basis. Intuitively, the bivector space specifies the rotation unambiguously (in 3-d, the plane and the oriented angle of rotation), and the final rotation is unambiguous, and thus also its matrix in terms of the original basis.