I hope he will forgive my making it public, but this entire thread was started at Chemboy's request. So if he wants to talk about groups, then so be it - he's the boss. I suggest the Lie groups as a compromise. But first me make something absolutely plain:

I make no claims that my attempts to explain manifolds is in any way thorough. Rather my intention was to give my friend, and other interested readers, just enough of a feel for the subject that, should they wish to dig deeper in textbooks (good Heavens - what are those?) or on sites like Wolfram or PlanetMath, they will at least have a fighting chance.

As DrRocket raid, it is a truly massive subject. But I hope we have gained some insight here. So will it be with the Lie groups. Warning: this will be devilishly long for a single post: I excuse it on the grounds it is mostly words, and a weekend.

Now this will be something of a digression, but you will see that maybe it's not too far afield if I offer a quick definition:

*A Lie group is a manifold endowed with the structure of a group*. The converse will serve equally as a definition, obviously.

My initial approach may raise a few eyebrows, but it has, I think, the advantage of being moderately intuitive. So we will start with a pair of garden-variety real vector spaces, and consider the

*totality* of linear maps (transformations, operators)

.

It is relatively easy to show that this collection of maps is itself a vector space over the reals (use

etc), but since this fact is at present of no interest to us, we may pass over it. Let's write this vector space as

.

By my notation, the space

has as elements (vectors) all the maps

. Of course, we introduce no ambiguity by writing this as

.

Now we know that any linear transformation a real space can be written as a matrix, whose entries are real; specifically, for a n-dimensional real space, these transformation matrices are real n x n matrices. We now want to restrict our attention to those maps which are invertible i.e. they are "isomorphisms", though here, the correct term is

*automorphism*.

In terms of the matrix representation of our automorphisms, this must mean that our matrices have non-zero determinant (this is because the formula for calculating the inverse of a matrix includes the factor

). By stipulating the existence of all inverses, and recognizing the identity matrix we will find that the vector space of all automorphisms, written as matrices, is in fact a group. This group we will call as

, the general linear group.

So the group

has matrix multiplication as the group operation, the identity

, say, and all inverse matrices such that

. Let us observe that, in general

.

Now we will ultimately be rewarded by concentrating, not on what groups

*do* but rather by what they

*are*. The obvious abstraction is to rewrite this group as

to signify this is

**the group of all real n x n matrices with non-zero determinant** (we call such a matrix "non-singular", btw).

This is an example of a Lie group, though to show this, we need to convince ourselves it is indeed a manifold.

So, let

be a group element, a matrix as defined above.

is an n x n matrix, i.e. there are

entries in

, which we will write as

, where i labels the rows, and j labels the columns, say. For ease of notation let's set

Then evidently there is a mapping

.

Now provided we are consistent in our mappings, say the i-th element in our m-tuple maps onto the jk-th entry in our matrix, we see that the the m-tuple

uniquely determines a matrix in

. Thus we have an isomorphism

.

If we can show continuity of this map, we will have a homeomorphism. Without attempting any degree of rigour, I offer this. Let

be the matrix with entries

. Then whenever the point

is arbitrarily close to

so the matrix whose entries are

is arbitrarily close the matrix whose entries are

.

This would imply a continuous isomorphism - a homeomorphism - so this group would indeed be a manifold.