Groups as symmetries, and question about Lie groups

Question

I am just starting out learning about group theory (so perhaps looking at Lie groups is a bit premature, but I am not hoping to have an in-depth understanding of them...yet) and when I looked up Lie groups I found the following statement on this page

Mathematicians invented the concept of a group to capture the essence of symmetry. The collection of symmetries of any object is a group, and every group is the symmetries of some object. E8 is a rather complicated group: it is the symmetries of a particular 57 dimensional object, and E8 itself is 248 dimensional!

I have two questions about this statement, which I think (hope!) are answerable to someone with just the basic knowledge in group theory.

• Firstly, I know there are many groups which are 'symmetries' of objects: permutation groups being the symmetries of some set, the dihedral groups, the Euclidean group etc

  However I thought groups were just... well, groups. And you could have groups which were symmetries. I don't quite understand how "every group is the symmetries of some object". I am thinking of a symmetry as being some mapping of an object onto itself that preserves some property. In the case of the Euclidean group the distance between two points is preserved, for dihedral groups the appearance of the n-gon is preserved, for the permutation group the collection of elements is preserved. But what about, for example, the GLn(F) groups of invertible matrices, or even the group of non-invertible matrices, or any of the many other groups. If a group is just a set with some operation on the elements, satisfying a set of axioms, how does this give rise to all groups being symmetries? Has it to do with the fact that the group must be closed?

• Secondly, the statement that "E8 itself is 248 dimensional" is not clear to me. I get how the group can be the symmetries of some geometric object (I assume that the E8 is the symmetries of a geometrical object-as opposed to some other mathematical object- and that is what the 'dimensions' are referring to') but if E8 is the symmetries of a geometrical object, surely it is not itself a geometrical object to have some dimensionality?! Either I am misunderstanding the term 'object' in this context, or the term 'dimension' in reference to the 57 dimensional object and 248 dimensional lie group E8 do not mean the same...

Answer 1

• Groups are "sets" of symmetries. This is a classical result known as the Cayley's theorem: every group is a subgroup of some symmetric group (possibly infinite). And the object that you are looking for (the one that always works) is, well, the group itself. :)

• surely it is not itself a geometrical object to have some dimensionality?! Why not? This is exactly what happens. There's an additional structure on $E8$. It is not only a group in the algebraic sense, but it is also a Lie group. Therefore it is a manifold so we can talk about its dimension. And it just happens to be 248.

Answer 2

Yes, it does seem premature to ask this question. To begin with, you have to learn a proper definition of a Lie group, otherwise, there is no way you can understand this statement about E8. First, make sure you have solid understanding of point-set (aka general) topology, at least for subsets of $R^n$. Then learn what differentiable manifolds are, Guillemin and Pollack "Differential Topology" is my favorite reference (as an introduction). You do not need to read the entire book, just get comfortable with definitions. Then learn what a Lie group is, say, by looking at one of the references collected in this MSE question. Or, at least, take a look at the wikipedia page. Now, at least you know meaning of the statement "E8 itself is 248 dimensional." If, after all this, you still want to know about "57-dimensional object", read in-depth discussion of partial flag-manifolds associated with simple (say, complex) algebraic Lie groups. The formal definition is that these are quotient manifolds $G/P$ where $P$ is one of the parabolic subgroups. You either work this out yourself, or ask at MSE about the codimension of the maximal parabolic subgroup of E8 associated with the highest root (the answer is 57). (Currently, I do not expect these words to mean anything to you.) Or just see p. 187 of

È. B. Vinberg, editor. Lie groups and Lie algebras, III, volume 41 of Encyclopaedia of Mathematical Sciences. Springer-Verlag, Berlin, 1994. Structure of Lie groups and Lie algebras.

Answer 3

(I)

Originally, there were symmetry groups: sets of symmetries of a mathematical object, like a figure in space or roots of a polynomial equation. A symmetry is a usually a transformation that preserves some chosen attributes of the mathematical object.

Then we distilled the abstract properties from symmetry groups to define an abstract group using axioms and without any reference to symmetry, only a binary operation and its properties. Thus we can study groups in their own right, and with the idea of isomorphism classes can compare the groups associated to different objects. (For example, a triangle has the same symmetry as a set of three elements, $S_3$, and a cube has the same rotational symmetry as a set of four elements, $S_4$.)

Going back to symmetry groups, one thing to notice about them is that there are often induced actions on various sets or structures associated to the original mathematical object. For example, given a symmetry group $G$ of a polyhedron in 3D space, there is an induced action of $G$ on the set $V$ of vertices, the set $E$ of edges, the set $F$ of faces, the set of space diagonals, the set of "flags," and so forth. This leads to the idea of group homomorphisms and group actions.

When every element of the group acts in a distinct way on a set $X$ (i.e. the map $G\to\mathrm{Perm}(X)$ is injective, in which case we call the action faithful), then $G$ is isomorphic to its image, so we can identify $G$ with a subgroup of $\mathrm{Perm}(X)$. Such a subgroup is called a permutation group. It is natural to ask, are "abstract groups" a more general class of groups than permutation groups? Could there abstract groups that are not isomorphic to permutation groups?

There aren't: every group acts faithfully on something. Let's say we have a group $G$ and want to describe an action of $G$ on some set $X$ but without assuming anything about $G$. First of all, we need a set to work with. We could pick a one-element set, but then $G$ acts trivially, which can't be a faithful action (unless $G$ is itself trivial of course). We could pick a two-element set, or some other set with a specific number of elements, but then there is no way to construct an action of $G$ on it without knowing anything about $G$ (indeed, depending on the sizes of $G$ and $X$, there may not be a nontrivial action at all!).

We have to create a set $X$ out of what we have available to us by hypothesis, but the only thing we have to work with is the group $G$ itself, so we simply use the regular action of $G$ on itself "by translation." That is, every $g\in G$ defines the permutation $\lambda_g:G\to G$ given by $\lambda_g(x)=gx$, and the map $g\mapsto\lambda_g$ is a homomorphism $G\to\mathrm{Perm}(G)$.

This is known as Cayley's Theorem. Often we interpret groups as symmetry groups, and symmetry groups can often look different (in particular, be "bigger") than the thing they are acting on, so describing the regular action as "translation" may in general be a bit unintuitive. However, the term "translation" makes perfect sense with the simplest examples: the real line $\mathbb{R}$, the circle $S^1$, or cyclic groups $C_n=\mathbb{Z}/n\mathbb{Z}$.

The real number line $\mathbb{R}$ may be identified with the group of translations of the number line. (In fact, this illustration works with $\mathbb{R}^2$, $\mathbb{R}^3$, etc.) And the circle $S^1$ may be identified with the group of 2D rotations. The cyclic group $C_n$ may be identified with the $n$th roots of unity, which form a $n$-gon in the circle group $S^1$ (itself viewed as a set of complex numbers), and this cyclic group is isomorphic to the group of 2D rotations by angles that are integer multiples of $2\pi /n$.

(II)

Yes, a symmetry group can itself be a geometrical object. Specifically, a Lie group is a smooth manifold. Indeed, $\mathbb{R}$ can be viewed as a symmetry group of itself (translations are the only transformations which preserve distances and do not reverse orientation); same with the circle group $S^1$ (rotations are the only transformations which preserve distances and orientation, again): both of these Lie groups are one-dimensional spaces.

Now let's go to two dimensions. Consider the group $\mathrm{Aff}(2,\mathbb{R})$ of orientation-preserving "affine transformations" of the plane $\mathbb{R}^2$. These are the transformations that preserve all distances (they are "isometries") and preserve orientation. Every such affine transformation is uniquely expressible as a translation and a rotation around the origin. There are a circle's worth of rotations around a point, which is $1$-dimensional, and there are an entire plane's with of translations, which is $2$-dimensional, for a total of $1+2=3$ dimensions! In fact, the space of all such affine maps of the plane is the same space as $S^1\times\mathbb{R}^2$! (The term is "diffeomorphic.")

(Note the boundaryless solid torus $S^1\times\mathbb{R}^2$ can itself be viewed as a Lie group, since $S^1$ and $\mathbb{R}$ are groups in their own right and it makes sense to take the direct product of groups. But while the groups $S^1\times\mathbb{R}^2$ and $\mathrm{Aff}(2,\mathbb{R})$ are the same space, they are not isomorphic as groups! One is abelian, the other isn't.)

Ultimately, "dimension" is a way to measure how many degrees of freedom there are, or for our purposes the number of real numbers needed in a parametrization. This is why it makes sense to add $1+2$ when determining the dimension of $\mathrm{Aff}(2,\mathbb{R})$ based on the decomposition into a knit product of $S^1$ and $\mathbb{R}^2$. Note these Lie groups are bona fide topological spaces with smooth structures, sometimes even metrics.

In particular, the 3D rotation group $\mathrm{SO}(3)$ may be viewed as a subset of the vector space of $3\times 3$ real matrices. Since this vector space has the a canonical norm (the Hilbert-Schmidt norm, associated to the Frobenius inner product in which the obvious canonical basis matrices are orthonormal), we can speak of the distance between points, and it turns out $\mathrm{SO}(3)$ is a $3$-dimensional submanifold of $M_3(\mathbb{R})$.

Here's a way to count the dimension. Every 3D rotation is a rotation around some axis by some angle. There are a 2D sphere's worth of oriented axes to choose from, and a circle's worth of angles to rotate by, for a total of $1+2=3$ dimensions. However, unlike with our previous example of $\mathrm{Aff}(2,\mathbb{R})$, it turns out $\mathrm{SO}(3)$ is not the same space as $S^2\times S^1$. Partly this is because a $0^{\circ}$ rotation around any axis is the same, and partly because rotation around an oriented axis by $\theta$ is the same as rotating around the opposite-pointing axis by the opposite angle $-\theta$.

However, it does turn out that $\mathrm{SO}(3)$ is "a bunch of circles arranged in the shape of a sphere" (the keyword here is "fiber bundle"), but they are in a sense twisted around the sphere. This is similar to how a Möbius band and a usual wristband can both be regarded as a bunch of line segments arranged in the shape of a circle, however the Möbius band has the line segment twist as it comes back around. But the twisting of circles around a sphere is harder to visualize.

(Also, it is probably easier to talk about the double cover $\mathrm{Spin}(3)$ of $\mathrm{SO}(3)$, since topologically it is just a $3$-sphere $S^3$ sitting inside four dimensions $\mathbb{R}^4$, and we have an associated "Hopf fibration" $S^1\to S^3\to S^2$.)

(In general, under nice conditions, there is a Lie group version of the "orbit-stabilizer theorem" in which if a group $G$ acts on a space $M$ with point stabilizer $S$, there is a fiber bundle $S\to G\to M$, which in particular implies $\dim G=\dim S+\dim M$. That applies here with $\mathrm{SO}(3)$ acting on $S^2$ with point stabilizer $\mathrm{SO}(2)\simeq S^1$.)

Note the symmetry group can be bigger-dimensional than the original space. The 3D rotation group $\mathrm{SO}(3)$ is $3$-dimensional and acts as the symmetry group of the $2$-dimensional sphere $S^2$ for example. Or if we consider the (orientation-preserving) affine transformations of 3D space instead of 2D, again every affine map is uniquely a translation and a rotation around the origin, for a total of $3+3=6$ dimensions, so in this sense the symmetry group of $\mathbb{R}^3$ is $6$-dimensional.

I said earlier that symmetry groups are comprised of symmetries, and symmetries are transformations that preserve a chosen set of attributes of a mathematical object. Well, we could ignore the metric / norm / inner product / topology etc. on $\mathbb{R}^3$ and only consider it as a vector space. In this case, the symmetries are all invertible linear maps, $\mathrm{GL}(3,\mathbb{R})$, again a subspace of the full vector space of matrices $M_3(\mathbb{R})$. In this case, $\mathrm{GL}(3,\mathbb{R})$ is in fact $9$-dimensional!