A topological group, or a continuous group, is a group G, with a topology placed upon it, such that group action and group inverse are continuous operators. A straightforward example is the reals under addition, using the linear topology. A nonabelian example is the rotations of the sphere in 3 space. Two rotations, combined in sequence, can differ by δ, and the composite rotation will differ by ε.
Remember how we prove continuity in general (G need not be a metric space). Take an arbitrary open set O, and an arbitrary point c in O, with inverse d, and find an open set U containing d, such that U inverse lies in O. Thus the preimage of O, under the inverse map, is covered with open sets, and is open, whence inverse is continuous. Group action is similar, but it is a binary operator, so you must use the product topology on G cross G. If xy = c, find an open set U containing x, and an open set V containing y, such that UV lies in O. That covers the preimage of O with open sets, and makes group action continuous.
The reals under addition, with the usual topology, forms a continuous group. In fact this is a continuous ring, since both addition and multiplication are continuous.
Any group can be made continuous; simply apply the discrete or indiscrete topology.
If M is an R module, and M and R each have a topology, and addition in M is continuous, and scaling by R is continuous, then M is a topological module, or a continuous module. If scaling is continuous then you don't have to prove the inverse group operator is continuous, since inverse = scaling by -1. When R is a field, such as the reals, M is a vector space, and M becomes a topological vector space, or a continuous vector space. However, the phrase topological vector space usually implies a real vector space, rather than some other field or division ring. If the field is not the reals then you must specify the field or division ring. Review the section on a real vector space as a continuous module, and then look at a general topological vector space. Translation by x, or scaling by nonzero c, is a homeomorphism on S, and carries any subspace to a homeomorphic image of that subspace. Also, a local base at 0 (or the identity if G is nonabelian), translated by x, for all x in G, becomes a base for the entire topology. Conversely, if the local base becomes the entire base, then continuity of addition at 0 translates to continuity everywhere. If U + V lies in W, all base open sets about 0, then x+U + y+V lies in x+y+U+V lies in x+y+W. Continuity of a nonabelian group must be proved for each xy.
If G is a continuous ring or module, scaling by a unit is a homeomorphism; but if c is not a unit, all bets are off. Let G be the integers mod 4, with the indiscrete topology. This is a continuous ring, yet scaling G (an open set) by 2 gives 0 and 2, which is not an open set in G.
Let H be a subgroup of G, and give it the subspace topology. Look at preimages of open sets in H, which come from open sets in G, and verify group action and group inverse are still continuous. Thus H is a continuous group. This applies to submodules and subrings as well.
Since translation by x is a homeomorphism, all cosets of a subgroup are homeomorphic to that subgroup. This means an open subgroup is closed, since the complement is a collection of open cosets of the subgroup. If a subgroup is closed and has finite index it is open, since the complement is finitely many closed cosets.
Any open subgroup of a compact group has finite index, since finitely many cosets cover G.
Let S be an open set in G. Let H be the subgroup or submodule generated by S. Select any a in S. If H contains x it contains y = x-a. Then H contains all of y+S (or y*S if G is nonabelian). This is a translate of S that includes x. H is covered with open sets and is open.
Assume H is a normal subgroup of G. Identify the points in each coset of H, thus giving the quotient group K = G/H. Give K the quotient space topology. Let a set S be open and consider the cosets of H represented by S. Given a in S times x in H, x is multiplied by everything in S, giving a translate of S. The represented cosets of H are covered by open translates of S, giving an open set. According to the subspace topology, this becomes open in K. Therefore open sets in G map to open sets in K, and the quotient map is bicontinuous.
Take an open set O in K, and let xy lie in O. Place x and y in open sets U and V such that U*V lies in the preimage of O. Push these down to K, and x and y lie in open sets whose product lies in O. Do a similar analysis for the inverse operator, and K becomes a continuous group. For H a submodule, K is a continuous module.
If H is open then every coset of H is open, every point in K is open, and K is discrete. Conversely, if K is discrete then 0 is open, and H (the preimage of 0) is open.
Similarly, H is closed iff every point of K is closed.
Let 0 → A → B → C → 0 be an exact sequence of topological groups. The arrows are morphisms in the category of topological groups, i.e. continuous group homomorphisms. As groups, B/A is isomorphic to C. We would like to show this is a homeomorphism. An open set in C pulls back to an open set in B by continuity. This becomes an open set in B/A by quotient correspondence. Thus the map from B/A onto C is continuous. If B is compact then the continuous image B/A is also compact. If C is hausdorff then the map from B/A onto C becomes a homeomorphism, and the spaces are equivalent.
This is a forward reference to algebraic topology; if you are not familiar with the fundamental group of a space you may want to skip this section for now.
The fundamental group of a continuous group G is abelian, even if G is nonabelian.
Let 1 be the identity of G. The fundamental group of G*G based at 1,1 is the direct product of π1(G) with itself. Let l1 and l2 be loops representing any two homotopy classes of G based at 1, and consider l3 = l1,1 + 1,l2 in G*G. Let m be the multiplication map from G*G onto G, which we know is continuous. Apply M to l3 and get l1+l2 in G. Apply m to a homotopy in G*G and find a homotopy in G. This is a consequence of the fact that m is continuous. A homotopy changes the first component of l3 from l1 + 1 to 1 + l1. This is merely a reparameterization of the loop. Instead of sitting at the base point for the second half of the unit interval, we sit at the base point for the first half. At the same time, a homotopy changes the second component of l3 from 1 + l2 to l2 + 1. Apply m, and l1+l2 is homotopic to l2+l1. Thus the fundamental group is abelian.
If f is the inverse function on G, and l is a loop in G based at 1, then l,f(l) in G*G maps, via m, to the constant loop at 1. This is homotopic to l,1 + 1,f(l) in G*G. The loop l, plus its inverse, becomes the identity in π(G). Inverse loops are inverse in the fundamental group.
You can use the same proof to show higher order π groups are abelian in G; although higher order π groups are abelian in any space, so this isn't new. However, the previous paragraph is new. One can apply f to the image of Sn to find the opposite homotopy class in πn(G).
A covering space of a continuous group is another continuous group, and the projection implements a continuous group epimorphism. This assumes G is path connected. If you aren't familiar with covering spaces you can skip this section.
Let's illustrate with the real line covering the circle. Let the circle have circumference 1, and place it tangent to the real line at 0. The real line wraps smoothly around the circle, and implements a group homomorphism from the reals under addition to the reals mod 1. Placing the circle at 0 was somewhat arbitrary. We could have placed the circle at 7, and 7 would become the identity for the group in the covering space. This gives results like 8+9 = 10, but it works. So if e is the group identity downstairs, a special point e~, in the fiber of e, is designated as the group identity upstairs. From there we create the group operation and prove it is continuous.
Let G be a topological group and let G~ cover G. Let e be the identity of G, and let e~ map to e under the projection p. Let m be the multiplication function downstairs, and define m~ upstairs as follows. Given a~ and b~ in G~, draw a path from e~,e~ to a~,b~ in G~*G~. (This is essentially two paths, one in each component.) Project this down to G*G, apply m, and lift this path back up to G~ starting at e~. The end of this path is m~(a~,b~) in G~.
Is this well defined? Let f be the projection of G~*G~ onto G*G, followed by m. Thus f is continuous from G~*G~ onto G. Group action is now f followed by a lift. Imagine some other path from e~,e~ to a~,b~. (This is once again a path in each component.) Combine this with the previous path to get a loop from e~,e~ back to itself. Apply f, and find a loop l in G. If l lifts to a loop in G~ then there is no trouble. Both paths lead to the same endpoint, the same m~(a~,b~).
Loops don't always lift to loops. Once around the circle lifts to a line segment in the real line, not a loop. But in this case l lifts to a loop.
As shown in the previous section, the loop upstairs is homotopic to the first loop passing through a~, crossed with e~, followed by e~ crossed with the second loop passing through b~. This homotopy projects down to a homotopy on l, and does not change the lift of l. Once this reparameterization has taken place, m has no effect. We multiply the first loop by e, then the second loop by e. We are left with the projection from G~ down to G of one loop, followed by the projection of the next loop. The lift is unique, hence the lift of the projection of the first loop is that same loop, and similarly for the second. The result is a loop in G~, whence group action is path independent and well defined.
Since any path will do, run a path from e~ to a~ in the first component G~, and combine this with the constant path e~ in the second component. Descend through f, and lift, and get the path to a~ back again. In the same way, e~*a~ = a~, and e~ is the group identity.
If i() is the inverse function in G, project a path from e~ to a~ down to G, apply i(), and lift, giving a path that ends in a point that I will call b~. Run the first path in the first component and the second path in the second component, from e~,e~ to a~,b~. Descend through f and find the constant path at e. Lift to find e~. Thus the inverse of a~ exists.
Consider a~b~c~ in G~. Reparameterize so that a path runs to a~ first, then to b~. Downstairs this is a path from e to a, then a path from a to ab. Lift, then apply f again in preparation for (a~b~)c~. The third path upstairs, from e~ to c~, becomes a path from ab to abc. Evaluate a~(b~c~) and find the same path in G, from e to a to ab to abc. The lifts are the same, and group action is associative.
Verify that G~ is abelian if G is abelian. Let both paths run together in time, and apply m, and it doesn't matter which path lives in which component.
The projection p is a function from one group onto another. Evaluate p(a~b~), and review the path downstairs which was lifted to produce a~b~ upstairs. The path ends at ab, hence p(a~b~) = ab. At the same time, p(a~) = a, and p(b~) = b, and this becomes ab in G. Therefore p is a group homomorphism. The kernel is the fiber of e.
Like any projection from a covering space to the base space, p is continuous.
The only thing left is the continuity of m~. You want to work inside an evenly covered neighborhood of ab, homeomorphic to a neighborhood about a~b~. Find open sets about a and b whose product lies in the neighborhood about ab, and restrict these to evenly covered neighborhoods, so that they lift to homeomorphic neighborhoods about a~ and b~. Within these neighborhoods, draw a path from a~,b~ to a nearby point in G~*G~. This path projects to the homeomorphic neighborhoods about a,b, and the product remains in the neighborhood about ab. Lift, and you are still in the neighborhood about a~b~. This establishes the continuity of m~. The continuity of the inverse operator is demonstrated in a similar fashion.
Put this all together and G~ is a continuous group, and p is a continuous group homomorphism onto G. In the same way, one can show G~ is a continuous ring if G is a continuous ring. In this case the projection p implements a ring homomorphism. Lift addition and multiplication up to G~ using paths and the corresponding operations in G. Verify all the properties of a ring. These ultimately come from the properties of the corresponding operators in G.
Let a filtration of the group G be a descending chain of normal subgroups of G. Now give G a topology whose open sets are the cosets of the normal subgroups in this filtration. Is this a base? Assume the cosets x1H1 and x2H2 intersect in z, where H2 lies in H1. The coset zH2 contains z, and equals x2H2, and lies in x1H1. The topology is valid.
Since all cosets are open, each base open set is closed.
Consider a coset xH and its preimage under multiplication, i.e. group action. Look at the product y1y2 = x in the factor group G/H, and lift to G. The cosets y1H and y2H are open, hence multiplication is continuous. Similar reasoning shows the inverse is continuous. G is a continuous group.
If the filtration ends with 1, or contains only G, we obtain the discrete or indiscrete topology for G, respectively.
A descending chain of ideals turns a ring R into a continuous ring, using the same proof as above.
Let M be a left R module and let Mi be HiM, where Hi is the ith ideal. Let cosets of Mi be open. This makes M an abelian continuous group. If cosets of H[i] are open in R, then M is a continuous module. Let cy lie in z+Mi. Evaluate c+Hi * y+Mi, and the product is still in z+Mi. That does the trick.
If 1 is the identity of G, let H be the intersection of all the open sets about 1, which is the same as the intersection of the local base.
If x lies in H then any open set O containing 1 has an inverse open set containing 1 and x, hence O contains 1/x. This puts 1/x in H.
Take x and y in H, and note that x,y in G*G lies in every base open set about 1,1. If xy is not in some neighborhood about 1, pull this neighborhood back by continuity of multiplication at 1. The preimage always contains x and y, hence our open neighborhood contains xy after all. Thus H is a subgroup of G.
With x in H, suppose yx/y is not in some open set O containing 1. This means x is not in 1/y*O*y, an open set containing 1. This is a contradiction, hence yx/y is in every open set containing 1, yH/y is in H, and H is a normal subgroup of G.
Suppose O is an open set that misses 1, yet O intersects H in y. O/y contains 1, and contains H. Thus O contains Hy. Since O does not contain 1, H cannot contain 1/y. This is a contradiction, hence every open set that misses 1 misses H entirely.
Let y be a point not in H; hence 1/y is also not in H. Let U be an open set containing 1 that does not contain y or 1/y. Let O be the open set Uy. O contains y, but not 1. Thus y belongs to an open set apart from H. This holds for each y outside of H, hence the complement of H is open, and H is closed.
Any open set that touches H contains 1, hence the closure of 1 is H. Nothing smaller than H (and still containing 1) can be closed. Points are closed iff H = 1, iff G is hausdorff, iff G is half hausdorff.
Given a coset of H and an open set O, O contains the coset, or misses the coset entirely. Translate the coset back to H to see this. As far as the topology is concerned, the points inside H, or any coset of H, are indistinguishable. The quotient group G/H uses the quotient space topology, but in this case an open set in G pushes down to an open set in G/H, which pulls back to the same open set in G. Each coset of H already belongs to the original open set in G, so nothing new is brought in. The quotient map G/H is bicontinuous. Since H is closed, the points of G/H are closed.
One may intersect the open sets about 0 in a topological ring and find a topologically closed additive subgroup H, as per the above, that is also closed under multiplication. See the xy ∈ H proof above. But remember, H need not be an ideal.
Let J be a subgroup of G, and let H be the closure of J. Employ the ab test to show H is a subgroup of G. Every open set containing a intersects J, and similarly bor b. Suppose a/b lies in an open set O missing J. Compose group action and inverse, and G/G is continuous. Place base open sets around a and b. Both sets contain elements of J, their product contains an element of J, and J intersects O. This is a contradiction, hence H is a subgroup of G.
As an example, the closure of the rationals is the reals, using the linear topology.
Let G be an abelian topological group written additively. A sequence xi is cauchy if for any neighborhood U about 0, there exists some n such that the difference between terms beyond xn lies in U. Let U be small balls about 0 in a metric space, and this definition agrees with the one you already know.
There is another common definition of cauchy: for some n, xi-xn lies in U for i ≥ n. Clearly xi-xn falls under the umbrella of xi-xj; hence the earlier definition implies this one. For the converse, find open sets V and W with V+W in U. This follows from the continuity of addition at 0. Let V = V intersect W, which includes 0, and is nonempty, whence V+V lies in U. Intersect V with -V, and now V±V lies in U. Let xi-xn lie in V, and consider xi - xj. This is (xi-xn) - (xj-xn), which is something in V - something in V, which lies in U. Thus the two formulations of cauchy are equivalent; we can use either one.
Two sequences are equivalent if their difference approaches 0, i.e. their difference can (eventually) be contained in any open set about 0. Verify this is an equivalence relation. Reflexive is obvious, and symmetric is easy; keep xi-yi in U by keeping yi-xi in -U. (Remember that group inverse is a homeomorphism.) Transitivity relies on the continuity of addition at 0. To keep Zi-xi in U, find any two open sets V and W with V+W in U. Keep zi-yi in V, and yi-xi in W, and you're all set. Thus sequences can be clumped together into equivalence classes.
Any sequence is equivalent to its tail. Let's take terms off one at a time. Beyond some n, differences between xi and xj are contained in U. This includes xi - xi-1. Thus x is equivalent to a shifted copy of itself, without the first term. Remove as many terms as you like; all that matters is the tail.
Build a new set G′, which is the equivalence classes of sequences as described above. This will be the completion of G. You probably know what we're going to do next. We want to prove, step by step, that G′ is a continuous group, with G as subgroup. The reasoning is almost identical to the completion of a metric space. With this as precedent, I will only outline the steps.
Let the sum of two sequences be the term by term sum. Prove the sum of two cauchy sequences is cauchy, using the U V W argument given above. This is where we need G to be abelian. (yj+xj) - (yi+xi) = yj-yi + xj-xi = something in V plus something in W, which lies in U.
Vary one of the summands by a sequence that approaches 0. The result is changed by a sequence that approaches 0, hence the result is the same. Addition is well defined on equivalence classes of sequences.
Addition is commutative and associative.
The zero sequence is the additive identity.
The opposite of a sequence x is found by negating each term. This plus the original equals 0. Thus G′ is an abelian group.
G maps into G′ via the constant sequences. Addition in G′ is defined by addition in G. This is a group homomorphism that becomes injective iff different sequences are never equivalent, iff every nonzero y in G misses an open set containing 0, iff the intersection of all the open sets about 0 is 0, iff 0 is closed in G, iff G is hausdorff.
The indiscrete topology turns G′ into 0. The discrete topology turns G′ into G.
If H is the intersection of the local base, any point in the sequence x can be replaced with another point from the same coset of H. Open sets cannot distinguish between these two. Thus points can be replaced with cosets of H in our convergent sequences. The completion of G is the same as the completion of G/H.
Assume G is a continuous left R module, whence scaling by any c in R is continuous. Let c act on a sequence by acting per term. If xi approaches 0, find an open set V such that cV maps into U. Restrict x to V, and cx lies in U. Therefore c*0 = 0 in G′ as expected. If y converges, restrict yi-yj to V, and cy has its differences in U. Thus cy is a valid sequence in G′. Furthermore, y may differ by e, a sequence that approaches 0, and c*(y+e) = cy + ce = cy + 0. The action of c is well defined. Verify the associative and distributive properties of a module, and G′ becomes a left R module.
It would be fun to place a topology on G′, but how? Perhaps a set of sequences is open if their tails all wind up in an open set O, but this causes trouble. Let xn = 1/n, and let O be the interval (0,1). It looks like x belongs, but 0, an equivalent sequence, does not belong. A point of G′ is and is not in O. Unless G has a norm, there isn't an obvious topology for G′.
Apply this to a filter group. The descending chain of subgroups is the local base. Assume G is hausdorff, or divide through by the intersection so that G becomes hausdorff. If the chain is finite, and the intersection is 0, then Hn = 0, 0 is open, and G is discrete. This isn't very interesting, so assume the chain is infinite. Still, it intersects in 0, so that G is hausdorff.
Complete G, and call the result G′. If x is a cauchy sequence, and H1 is the first subgroup, xi-xn eventually lives within H1 for i ≥ n. This puts the tail in a particular coset of H1, determined by xn. Move on to H2, H3, etc. Build a sequence of cosets Hi, each holding the tail of x. (This assumes the descending chain of subgroups is countable.) Each coset lives in the coset that came before. The result is a sequence of cosets of Hi in G, or if you prefer, a sequence of cosreps drawn from G/Hi such that each is consistent with the previous. I call this a coherent sequence.
If x and y produce the same coherent sequence they are equivalent. For each Hn, their tails fall into the same coset of Hn, and x - y converges to 0. Conversely, each coherent sequence builds a convergent sequence in G′. Let xn be any element in the designated coset of Hn, and you're there. Finally, different coherent sequences are not equivalent, as they wind up in different cosets of Hn for some n. Thus we have characterized G′.
Addition takes place per term in the coherent sequences. Each sum of terms lies in the coset determined by the sum of the previous terms. The result is coherent as expected.
If G is a continuous module, and H is a chain of submodules, scaling takes place per term in the corresponding coherent sequence.
If you specify, in advance, coset representatives for all the cosets of Hn+1 in Hn, you can build a coherent series instead of a coherent sequence. Start with the coherent sequence, then replace the first coset with its predetermined cosrep of H1 in G. Subtract this from the rest of the sequence. This translates everything back into H1. Move to the next element and find a coset of H2 in H1. Replace this with the corresponding cosrep, and subtract this cosrep from everything else. Repeat this for H3 in H2, H4 in H3, and so on. The result is called a coherent series, because the partial sums reproduce the coherent sequence. Once again, elements are added term by term, but there may be some carry operations. Review the p-adic numbers.
The completion can be given a corresponding topology. Coherent sequences that agree on the first n terms form an open set. These are the sequences of G′ that wind up in a particular coset of Hn. This is a base for the topology of G′, just as it is a base for the topology of G. The local base is determined by the subgroups Hn. Use this to filter G′, and obtain the aforementioned topology. Take the completion, and get the same coherent sequences back again. The completion of G is complete.
If f(G) is a continuous group homomorphism into another topological group, the image of a cauchy sequence remains cauchy. If image points cannot be trapped in an open set then the original sequence cannot be trapped in the preimage of that open set. Thus f induces a group homomorphism on cauchy sequences. Completion becomes a functor from continuous groups into groups, or from continuous modules into modules. Restrict the first category to filter groups, and the second category becomes complete filter groups.
Recall that one can use open circles or open squares as a base for the topology in the plane; the resulting space is the same. In other words, open sets map to open sets. If G is a group, we are going to construct two filtrations that lead to the same completion with the same topology. To be more general, let's consider an R module M. Each filtration will then be a descending chain of submodules.
Let Fi and Gi be two filtrations on the module M, with nonnegative integers a and b such that Fa+i is contained in Gi, and Gb+i is contained in Fi for all i ≥ 0. (F0 and G0 are equal to M.) The two filtrations have "bounded difference", and produce the same completion.
Let x be a cauchy sequence through F. For every i, the differences between the terms of x eventually lie in Fa+i, which lies in Gi, hence x is cauchy through G. Furthermore, if x and y are equivalent, then y-x lies in Fa+i, which lies in Gi, and the two sequences are still equivalent. Sequences in the completion through F map to sequences in the completion through G, and by symmetry, this map can be reversed. This builds a bijection between the two sets.
Since each cauchy sequence maps to itself under a new topology, the map respects addition and scaling by R. The two modules are isomorphic.
A base open set in the first completion is the collection of sequences that wind up in some coset of Fi for some i. Let x be one of these sequences. Remember that x is cauchy through G, so it winds up in a coset of Gb+i. This coset is contained entirely in Fi. A base open set with respect to G, containing x, lies entirely in our base open set with respect to F. This establishes continuity. By symmetry, the completions are homeomorphic. The identity map on sequences becomes the module isomorphism and the homeomorphism.
The bounds need not be simple linear functions of i; any increasing function of i will do. For instance, a+i might be replaced with 2i2+7i+3. Run the above proof using a(i) and b(i), instead of a+i and b+i.
If H is an ideal of R, filter M by the powers of H, and then by the powers of H2. These are equivalent. Set a(i) = 2i, and b(i) = i.
A stable filtration on a module M with respect to an ideal H in the ring R is a filtration F with Fi+1 containing H*Fi, and Fi+1 = H*Fi for i > a. In other words, containment becomes equality beyond level a.
Set i = 0, and F contains HM. Multiply by H and HF contains HHM. This means F2 contains HHM. By induction, Fi contains HiM.
Clearly M contains Fa. Multiply by H and HM contains HFa. The latter is Fa+1. By induction, HiM contains Fa+i. Combine this with the previous paragraph, and the two filtrations have bounded difference, and give the same completion. Every H stable filtration gives the same completion as Hi times M.
Let P be a submodule of M, and filter M by the powers of an ideal H. Restrict attention to the set of convergent sequences with elements taken from P, and give this set the subspace topology. Compare this to the completion of P by the powers of H. Clearly Hi*P is contained in Hi*M intersect P. The other direction is trickier. Let Q be the submodule of M whose elements are eventually driven into P by some power of H. Assume Q is finitely generated. This is assured if M is noetherian, or finitely generated over a noetherian Ring. Thus some Hb drives Q into P, and Hb drives all the elements of M into P that are ever going to wind up in P. Hb+i*M (restricted to P) lies in Hi*P. The two filtrations have bounded difference, and the completions are the same.
A similar argument shows the completion of M/P (through the powers of H) is M′ mod (M′∩P). The latter module is the same as P′. (M/P)′ is convergent sequences of cosets of P in M, and M′/P′ takes sequences from M mod sequences from P, which are, term by term, sequences of cosets of P in M. The spaces are the same.
For a nonnoetherian example, let P = Z/25, (the integers mod 25), and let M be the direct limit of Z/5, living as multiples of 5 inside Z/25, living inside Z/125, living inside Z/625, and so on forever. This is a chain of increasing submodules, hence M is not noetherian. M is a Z module; let H = 5Z. If x is in M it is in a ring mod 5k for some k. A sufficiently high power of H drives x into P, but no power of H drives all of M into P. Let's compare the completions. Since HM = M, the completion through H is 0. Intersect with P and get 0. However, if we restrict to P, and complete through H, H2P = 0, hence P is discrete, and P′ = P.
Another example sets P equal to the polynomials K[x], M is the rational functions K(x), and H is multiplication by x. Again, HM = M, so that completion is 0. Complete K[x] by its maximal ideal x and get the power series K[[x]].