WEBVTT
Kind: captions
Language: en
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Now, I am back; and I had promised you last
time that we shall discuss subgroups of SO
00:00:24.520 --> 00:00:37.230
3. So, what are all subgroups of SO 3, when
I say SO 3, I mean SO 3 R which is also group
00:00:37.230 --> 00:00:51.990
of rotations in R 3. And I had emphasized
it towards end of last lecture that all platonic
00:00:51.990 --> 00:00:58.950
solids can be inscribed in this sphere, everything
is perfectly symmetric. So, I can perfectly
00:00:58.950 --> 00:01:08.490
put them inside sphere; all all all these
one of them I can put them inside this sphere,
00:01:08.490 --> 00:01:14.180
yeah .
So, if I put them inside this sphere, and
00:01:14.180 --> 00:01:22.010
I do all the operations which are symmetric
for these objects, I am actually rotating
00:01:22.010 --> 00:01:28.530
the sphere that is I am considering, therefore
these symmetric operations of these objects
00:01:28.530 --> 00:01:39.230
as elements of SO 3, so that is easy . Therefore,
conclusion is that all those elements which
00:01:39.230 --> 00:01:51.650
were groups of rotations of tetrahedron
you remember what it was, it was A 4 rotations
00:01:51.650 --> 00:02:13.510
of cube that was A 5, rotations of dodecahedron
or icosahedron that was sorry you rotations
00:02:13.510 --> 00:02:29.180
of cube is
00:02:29.180 --> 00:02:36.500
rotations of cube is S 4 and rotations
of dodecahedron A 5.
00:02:36.500 --> 00:02:43.110
So, from the fact from the observation that
all platonic solids can be inscribed in a
00:02:43.110 --> 00:02:55.299
sphere, what we conclude is that all these
are sub groups of they are sub groups of SO
00:02:55.299 --> 00:03:06.419
3. And we had also observed last time that
cyclic groups which are any ways subgroups
00:03:06.419 --> 00:03:17.389
of SO 2 are also subgroups of SO 3, say finite
cyclic groups if you wish finite cyclic groups.
00:03:17.389 --> 00:03:30.169
So, secondly, if you want to list finite subgroups
of SO 3, you will have to have A 4, A 5, S
00:03:30.169 --> 00:03:44.120
4 and all cyclic groups. But the question
is what are all finite subgroups of SO 3.
00:03:44.120 --> 00:03:50.749
So, these are secondly few but what about
others and that is where the fine part is
00:03:50.749 --> 00:03:58.090
going to be and we are ready with group action.
So, it is a good application of group actions.
00:03:58.090 --> 00:04:05.719
Let me read out the theorem for you. If you
have a finite subgroups of SO 3, then following
00:04:05.719 --> 00:04:13.260
are the options for it nothing else can be
finite subgroups of SO 3. It could be finite
00:04:13.260 --> 00:04:27.980
cyclic group, finite dihedral group, Klein's
4-group or these three things which are coming
00:04:27.980 --> 00:04:44.510
from platonic solids . And this we have already
observed; this also we have observed.
00:04:44.510 --> 00:04:52.930
So, this we have to observe today, but more
than that we are suppose to show that if you
00:04:52.930 --> 00:05:01.000
pick any finite group finite subgroup of SO
3, then that subgroup is one of them, there
00:05:01.000 --> 00:05:08.660
is no other, so nothing which is not there
in this list is going to be a finite subgroup
00:05:08.660 --> 00:05:16.300
of SO 3, so that is quite amazing. So, as
I mentioned C n appears because of SO 2, and
00:05:16.300 --> 00:05:22.350
D n should appear. And D n appears because
regular polygons the two-dimensional ones
00:05:22.350 --> 00:05:27.260
all regular polygons can be inscribed in this
sphere that is correct. You have this sphere,
00:05:27.260 --> 00:05:33.220
you consider the equator of this sphere that
is circle and on that circle you can have
00:05:33.220 --> 00:05:37.740
all regular polygons.
So, if you have regular n 1, what you have
00:05:37.740 --> 00:05:44.870
is the dihedral group right. Because you
can have just imagine as if this is circle,
00:05:44.870 --> 00:05:53.020
when I move, I get cyclic group; and when
I rotate it along some axis on this sphere,
00:05:53.020 --> 00:05:58.680
what I am doing this actually I am flipping
the plane of the equator. So, rotating in
00:05:58.680 --> 00:06:05.160
three dimension like this that is swiping
northern south pole via rotations is actually
00:06:05.160 --> 00:06:14.720
same as flipping of the of the regular
r-gon which is there on the equator, so that
00:06:14.720 --> 00:06:24.430
is how you get the group D n. So, D n is fine.
What about Klein's 4-group, how do we imagine
00:06:24.430 --> 00:06:33.000
Klein's 4-group we are going to see after
few slides. But remember the purpose is we
00:06:33.000 --> 00:06:37.810
shall start with arbitrary finite group which
is subgroup of SO 3, and then we shall show
00:06:37.810 --> 00:06:43.960
that it has to be one of these, one of these
which is there in the list ok.
00:06:43.960 --> 00:06:55.810
So, for this theorem, I am going to have proof.
So, I have to start with arbitrary finite
00:06:55.810 --> 00:07:05.150
subgroup of SO 3 . So, I start with a finite
subgroup of SO 3. Now, what is there to start
00:07:05.150 --> 00:07:12.490
with when we can have what are called poles,
we can have poles of rotations. So, what is
00:07:12.490 --> 00:07:20.050
that? I will just take one of these objects.
So, I take this and I imagine that I have
00:07:20.050 --> 00:07:27.050
inscribed it inside a sphere. So, this these
vertices will be touching this sphere right,
00:07:27.050 --> 00:07:31.820
phases will be slightly away from them only
the vertices will be touching this sphere
00:07:31.820 --> 00:07:37.520
that is what is a meaning of inscribed. So,
all those points where these vertices are
00:07:37.520 --> 00:07:45.190
touching, they are potential axis for symmetry
right, I just hold these vertices and I just
00:07:45.190 --> 00:07:49.830
rotate.
So, points which are touching this sphere
00:07:49.830 --> 00:08:01.190
these vertices, so opposite vertices are forming
potential axis for the for a symmetry operations.
00:08:01.190 --> 00:08:08.010
And these things are going to be called poles
something more. I can have rotation through
00:08:08.010 --> 00:08:17.830
an axis which is the opposite sides, opposite
phases right. So, if you just extend them,
00:08:17.830 --> 00:08:23.680
extend them to touch this sphere, similarly
extend the bottom you touch this sphere, then
00:08:23.680 --> 00:08:28.340
the point where you touches this sphere is
again going to be called a pole. And similarly
00:08:28.340 --> 00:08:32.979
for your edges. So, all these things are going
to be poles for this object.
00:08:32.979 --> 00:08:41.959
So, I start with some considerations on poles.
So, what I have what I have been given is
00:08:41.959 --> 00:08:49.439
just arbitrary finite subgroup of SO 3, which
apriori has know geometry. So, I cannot really
00:08:49.439 --> 00:08:56.399
express the poles in terms of what I did,
but I will do it in some other terms. One
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thing is very clear each elements of G is
the rotations it is an elements of SO 3 . So,
00:09:06.170 --> 00:09:17.870
this is every elements the rotation, there
is an axis . Every rotation has an axis, axis
00:09:17.870 --> 00:09:26.079
of rotation. So, I denote by l g the axis
of rotation and then I intersect axis of rotation
00:09:26.079 --> 00:09:37.100
with S 2, S 2 is this sphere . So, what do
I get that is how I get poles. So, l g just
00:09:37.100 --> 00:09:49.300
take it to be infinite line, infinite length
line. So, all the axis all the way is going
00:09:49.300 --> 00:09:56.730
to intersect with S 2. And those two points
where intersecting with S 2 is those points
00:09:56.730 --> 00:10:02.740
are going to be called poles.
So, it is a definition elements of X you call
00:10:02.740 --> 00:10:18.839
them poles . And l g intersection S 2, these
are two elements, the two elements . This
00:10:18.839 --> 00:10:22.910
this intersection one just say North Pole,
other you just say south poles. So, they are
00:10:22.910 --> 00:10:26.260
called poles of g, poles for that particular
axis.
00:10:26.260 --> 00:10:36.029
And it is clear that since G is finite for
each G you have one l g. So, there are finitely
00:10:36.029 --> 00:10:43.019
many poles, number of poles is finite. And
if G is cyclic group, then there are only
00:10:43.019 --> 00:10:50.450
two antipodal points in the pole right, because
when you have some cyclic group again I am
00:10:50.450 --> 00:10:56.399
taking this cyclic just imagine as this is
sphere. When I am having cyclic group, there
00:10:56.399 --> 00:11:05.519
is only one pole, there is only one axis . Cyclic
group meaning is only one axis . Therefore,
00:11:05.519 --> 00:11:13.790
X will have only two elements just the antipodal
points.
00:11:13.790 --> 00:11:23.670
Now, what is the what is the idea of the
proof. Remember, we are proving there are
00:11:23.670 --> 00:11:33.199
finitely many possibilities for subgroup of
SO 3 . We have defined these poles; we shall
00:11:33.199 --> 00:11:38.920
do some juggler with those poles. We shall
make some observations about how those poles
00:11:38.920 --> 00:11:45.449
are distributed all across this sphere. And
depending on our observations of the distribution
00:11:45.449 --> 00:11:51.120
of poles all across the sphere, we are going
to make predictions about what G could be;
00:11:51.120 --> 00:11:57.279
in fact, that distribution would completely
be determining what our group is ok.
00:11:57.279 --> 00:12:07.209
So, it is first observations. First observation
is that one can give a very nice action of
00:12:07.209 --> 00:12:17.209
G on X. So, what your G, we started with it
actually acts on set of poles of it, poles
00:12:17.209 --> 00:12:27.309
of G which in the set of poles is being
denoted by X. It is easy to see let see. So,
00:12:27.309 --> 00:12:39.779
I take some elements which is a pole. And
pole for axis for some h. So, h is eventually
00:12:39.779 --> 00:12:50.480
an element of SO 3; and for that SO 3 element
and I have axis and then I have pole.
00:12:50.480 --> 00:13:01.459
So, what is clear is that, h this rotation
does not move this pole that is correct. If
00:13:01.459 --> 00:13:07.339
I am rotating like this, then north pole and
south pole they do not move, so h P is P.
00:13:07.339 --> 00:13:15.380
And now if I take any arbitrary element in
G in G, then I mean this calculation g h g
00:13:15.380 --> 00:13:26.800
inverse acting on g P, g P is what happens
to this pole and influence of g. So, g h g
00:13:26.800 --> 00:13:38.129
inverse g P is simply g h P, which is g h
P, h P is not moving P because h is precisely
00:13:38.129 --> 00:13:43.610
the rotation for is pole. So, I have this
to be g P.
00:13:43.610 --> 00:13:59.870
So, what is happening therefore g h g inverse
g P is g P that means there is some element
00:13:59.870 --> 00:14:09.449
in G, which is not able to move g P, that
means, this element g P is fixed by g h g
00:14:09.449 --> 00:14:14.879
inverse, and therefore it is fixed under the
influence of some element some rotation because
00:14:14.879 --> 00:14:21.540
every element of G is a rotation eventually.
So, therefore, g P is also a pole and for
00:14:21.540 --> 00:14:30.509
this pole, the axis is l g h g inverse, so
that particular axis is having g P as a pole.
00:14:30.509 --> 00:14:40.820
So, what we have concluded if P is a pole,
then g P is also a pole. What is the action,
00:14:40.820 --> 00:14:47.070
it is just simply that g as a rotation is
trying to move P that is a rotation. So, g
00:14:47.070 --> 00:15:21.209
P is influence of rotation determined by P
on sorry determined by g
00:15:21.209 --> 00:15:40.980
on P so that is an action G acts on poles
. And as I said we are going to investigate
00:15:40.980 --> 00:15:48.209
distribution of poles, and what are the orbits
of that action all that is what we are going
00:15:48.209 --> 00:15:52.679
to analyse.
So, we start with you remember Burnside's
00:15:52.679 --> 00:15:58.970
formula in very beginning I had mentioned
Burnside's formula, what it is number of orbits
00:15:58.970 --> 00:16:05.709
is equal to average number of fixed points.
So, this is this number of orbit for this
00:16:05.709 --> 00:16:16.499
action, this number of orbits for the action
in quotient is this . And why this is there,
00:16:16.499 --> 00:16:25.059
so for identity, identity is going to fix
all the poles how many poles are there, size
00:16:25.059 --> 00:16:34.230
of the set X, X is precisely the set of poles.
And what is this rest of the mod g minus 1
00:16:34.230 --> 00:16:42.339
non identity points they are fixing only two
poles which are on the axis which are the
00:16:42.339 --> 00:16:47.062
two sides of two edges of two nodes of the
axis.
00:16:47.062 --> 00:16:59.829
So, those are mod G minus 1. So, these mod
x is precisely size of fixed points of identity
00:16:59.829 --> 00:17:15.660
and for non identity. So, if g is different
from identity, then fixed points are going
00:17:15.660 --> 00:17:24.550
to be only two right, so that is how I get
this . Again and again quite frequently I
00:17:24.550 --> 00:17:32.840
am going to use this formula.
Now, I am just writing this in slightly different
00:17:32.840 --> 00:17:39.210
way. So, for each orbit, I choose representatives
x 1, x 2, x n. So, from each orbit, I am picking
00:17:39.210 --> 00:17:46.060
one element, and then I rewrite the summation
in this form. So, everything is fine here
00:17:46.060 --> 00:17:54.610
the only thing is in case of x I am just writing
this . So, for each so I am having for each
00:17:54.610 --> 00:18:00.130
orbit, one element and that element is contributing
how much equal to the size of its orbits.
00:18:00.130 --> 00:18:08.310
So, this is precisely the count of all orbits
and union of all orbits is precisely your
00:18:08.310 --> 00:18:14.210
set on which this group is acting. So, you
have this .
00:18:14.210 --> 00:18:22.410
And in fact some manipulations I would do
with this . So, I take this to the other
00:18:22.410 --> 00:18:32.050
side and divide by G. So, what I get are these
quantities . So, this is n minus 1 by summation
00:18:32.050 --> 00:18:41.981
G summation of all the orbits. And here I
I write orbit x i and take mod G inside. And
00:18:41.981 --> 00:18:52.270
you remember orbits stabilizer formula . So,
orbit of an element divided by orbit of G
00:18:52.270 --> 00:19:01.050
is 1 divided by stabilizer of x i because
orbit into stabilizer size, size of orbit
00:19:01.050 --> 00:19:07.250
times size of stabilizer equals size of the
group. So, so what we get after all these
00:19:07.250 --> 00:19:15.100
manipulations is the 2 times 1 minus 1 divided
by order of group is equal to 1 minus 1 divided
00:19:15.100 --> 00:19:22.140
by size of stabilizers and take the summation.
So, this formula again I am going to use multiple
00:19:22.140 --> 00:19:26.730
times, this also I am going to use multiple
times.
00:19:26.730 --> 00:19:34.320
If group is non-trivial that is group size
is more than better than equal to 2. So, I
00:19:34.320 --> 00:19:41.920
have this. This because group is non-trivial.
So, here I have 1 by n, where n is greater
00:19:41.920 --> 00:19:49.460
than or equal to 2. So, this number is certainly
more than 1. And since this is less than 1,
00:19:49.460 --> 00:19:58.150
twice of something less than 1 is less than
2. So, this is what I have . And this expression
00:19:58.150 --> 00:20:05.770
2 times 1 minus 1 by G from this is summation
of 1 minus 1 divided by size of stabilizers.
00:20:05.770 --> 00:20:13.580
So, just that much I replaced here, so 1 by
2 here, 2 here. So, I get this expression.
00:20:13.580 --> 00:20:19.120
And now I am taking summation overall orbits,
this is a constant half number of orbits is
00:20:19.120 --> 00:20:27.831
n, constant 1 number of orbits is n, and here
the summation is happening right. So, what
00:20:27.831 --> 00:20:34.520
I have is n by 2 is less than equal to twice
of this because this is precisely the summation
00:20:34.520 --> 00:20:40.400
is precisely twice of 1 minus 1 divided by
size of G, so I get this expression.
00:20:40.400 --> 00:20:47.290
So, n by 2 is less than equal to this is what
I get. And this expression is any way less
00:20:47.290 --> 00:20:57.100
than 2. So, when I combine this, this is strictly
less than 2; and therefore, m which is number
00:20:57.100 --> 00:21:04.400
of orbits is strictly less than 4. And also
by group as size greater than equal to
00:21:04.400 --> 00:21:11.460
2, so when I take that into considerations.
And look at this when I look at this what
00:21:11.460 --> 00:21:21.330
I get is half is less than 1 minus 1 divided
by mod G less than n by 2. So, again this
00:21:21.330 --> 00:21:29.370
is going to be quite useful for me.
So, from this what I conclude is that n is
00:21:29.370 --> 00:21:37.930
greater than 1. So, n is greater than 1 and
n is less than 4, that means there are only
00:21:37.930 --> 00:21:45.310
two possibilities that whenever group sub
finite sub group of SO 3 acts on its poles,
00:21:45.310 --> 00:21:52.640
number of orbits is either 2 or 3 that is
quite substantial information. And from
00:21:52.640 --> 00:21:58.180
this information, we are going to get all
possible subgroups of SO 3. And in fact, we
00:21:58.180 --> 00:22:03.150
are going to show that they essentially come
from catering solids or polygons, which
00:22:03.150 --> 00:22:11.400
is in case of SO 2. So, case by case analysis
first we do for n is equal to 2 case; and
00:22:11.400 --> 00:22:14.750
then we do analysis for n is equal to 3 case
.
00:22:14.750 --> 00:22:26.090
So, n is equal to 2 case . There are only
two orbits. So, if n is equals to 2, then
00:22:26.090 --> 00:22:42.240
size of x is 2. Why is that? Let us go back
and look at this . So, if n is 2, then 1 divided
00:22:42.240 --> 00:23:01.000
by size of G times mod X plus size of G minus
1 times 2 . So, what happens here in order
00:23:01.000 --> 00:23:15.790
to get 2 here, so let me just get mod G that
side. So, twice of size of G is therefore,
00:23:15.790 --> 00:23:24.220
size of X plus twice of size of G minus 2.
So, this cancels out and you get size of X
00:23:24.220 --> 00:23:36.010
is 2. So, size of X is 2, if n is 2.
So, number of poles is only 2. And if there
00:23:36.010 --> 00:23:46.750
are only 2 poles they have to be antipodal.
So, they have to be antipodal . And both of
00:23:46.750 --> 00:23:52.870
them are in different orbits, because the
number of orbits is 2. So, I cannot move this
00:23:52.870 --> 00:24:00.110
pole to that pole right. So, what is the picture
now that these are actually antipodal, I cannot
00:24:00.110 --> 00:24:07.820
move this from that and this is happening
when x 1, x 2 the line joining those two poles
00:24:07.820 --> 00:24:14.070
is actually the axis. And now your group is
moving and this is rigid the only possibility
00:24:14.070 --> 00:24:22.210
is your group is cyclic. So, all cyclic groups
occurs as subgroup of SO 3, and n equals to
00:24:22.210 --> 00:24:28.570
2 possibility is that one that was quite straightforward,
n to be precisely cyclic groups.
00:24:28.570 --> 00:24:34.220
Now, n is going to 3 cases something quite
something which is interesting. So, can you
00:24:34.220 --> 00:24:41.220
imagine what n is going to 3 would be a corresponding
to, when you think of platonic solids like
00:24:41.220 --> 00:24:50.640
this say like this. There are three types
of poles, poles which are coming from vertices,
00:24:50.640 --> 00:24:55.980
poles which are coming from edges and poles
which are coming from mid points of phases.
00:24:55.980 --> 00:25:01.850
And pole which is coming from midpoint of
phases cannot go to for that it cannot go
00:25:01.850 --> 00:25:09.150
to the pole which is on the midpoint of edge
or a vertex that is the interpretation. So,
00:25:09.150 --> 00:25:20.460
this 3 stands for, 3 stands for edges, phases
and vertices, axis determined by them, axis
00:25:20.460 --> 00:25:32.510
determined by them. And for that matter every
vertex is as good as any other vertex.
00:25:32.510 --> 00:25:39.300
As good as in what sense they are in the same
orbit. Similarly, any midpoint of edge
00:25:39.300 --> 00:25:46.720
is as good as any other, because they are
in same orbit by certain rotation, by certain
00:25:46.720 --> 00:25:56.130
action of the group of symmetry of this,
you can obtain one from the other ok . So,
00:25:56.130 --> 00:26:05.220
n is 3. So, what we do we take representatives
from these three orbits, and we are calling
00:26:05.220 --> 00:26:09.830
them are your x 1, x 2, x 3, x n let us simply
call them x, y, z.
00:26:09.830 --> 00:26:17.870
And one of the formulas that we had earlier,
remember what formula it was, summation of
00:26:17.870 --> 00:26:29.170
stabilizers . This formula, this formula,
this formula which says that twice of 1
00:26:29.170 --> 00:26:40.260
minus 1 by this is summation of stabilizers.
So, using that you conclude this thing that
00:26:40.260 --> 00:26:48.910
1 is less than 1 plus 2 divided by G is precisely
some of reciprocal of sizes of stabilizers.
00:26:48.910 --> 00:26:57.140
Now, here comes an observation. How large
could these be, how large these could these
00:26:57.140 --> 00:27:04.350
be. If all of them are greater than 1 by 3,
then this quantity will be less than 1, but
00:27:04.350 --> 00:27:09.600
here it is actually more than 1. So, what
we are therefore force to this that one of
00:27:09.600 --> 00:27:17.870
the elements say x, one of the elements representatives
of orbits has to have its stabilizer to be
00:27:17.870 --> 00:27:24.360
of order 2; otherwise we cannot have this
inequality. So, in order to respect this inequality,
00:27:24.360 --> 00:27:30.850
one has to have size of stabilize of one of
the orbits representatives to be equal to
00:27:30.850 --> 00:27:35.770
2.
What does it mean, it means that edges are
00:27:35.770 --> 00:27:46.250
there. So, 2 corresponds to edges. So, x here
therefore is corresponding to poles which
00:27:46.250 --> 00:28:00.770
are coming from edges . So, x corresponds
to poles coming from edges , but we do not
00:28:00.770 --> 00:28:09.611
need all this these proceed I am just making
size remark. Now, what are other possibilities,
00:28:09.611 --> 00:28:14.640
other possibilities are precisely these. So,
I have written all all possibilities. So,
00:28:14.640 --> 00:28:22.950
stabilizer 2, then next element other orbit
has stabilizer size 2, and other one has says
00:28:22.950 --> 00:28:33.690
stabilize size r, for some r and similarly
3, 3, 5, 3, 4, 5 like
00:28:33.690 --> 00:28:35.940
that ok.
Consider this situation. When n is 3 stabilizer
00:28:35.940 --> 00:28:42.500
of axis of size 2, stabilizer of y is also
size 2, and stabilizer of z is of size say
00:28:42.500 --> 00:28:54.650
r, then what happens? So, if r is equals to
2, r is this . So, stabilizer of z is of is
00:28:54.650 --> 00:29:04.830
also of size 2, then in one of the earlier
formulas as if we put all these values which
00:29:04.830 --> 00:29:11.570
formula, this one. So, here everything is
2, 2, 2. So, 1 by 2 plus 1 by 2 plus 1 by
00:29:11.570 --> 00:29:18.040
2 is 1 plus 2 by mod G. And that gives us
that order of G is 4.
00:29:18.040 --> 00:29:25.320
Now, as you know there are only 2 groups of
order 4, one is cyclic group, and another
00:29:25.320 --> 00:29:33.770
one is Klein's 4-group . Cyclic group is
certainly there in SO 3. What about Klein's
00:29:33.770 --> 00:29:40.390
4-group? Klein's 4-group is also there in
SO 3. How to see that and that is quite interesting
00:29:40.390 --> 00:29:47.790
description. We think of it is sphere and
you think of three axis x-axis, y-axis and
00:29:47.790 --> 00:29:53.610
z-axis, so that forms a system, that forms
a orthonormal system and consider with x axis,
00:29:53.610 --> 00:29:59.130
the minus direction as well with z axis, minus
direction as well and with y axis also minus
00:29:59.130 --> 00:30:09.360
directions. So, the six the you just consider
imagine the three vertex three axis intersecting
00:30:09.360 --> 00:30:16.710
at the centre intersecting at the origin.
Now, what is the symmetry, what are the symmetries
00:30:16.710 --> 00:30:22.990
of this. So, you consider rotation by 180
degree for each of them. So, you imagine rotations
00:30:22.990 --> 00:30:31.280
by 180 degree about this standard x, y and
z axis . So, then these rotations along with
00:30:31.280 --> 00:30:39.290
the identity, they form a subgroup of SO 3,
which is isomorphic to V 4. In fact, here
00:30:39.290 --> 00:30:48.200
you can think in terms of quaternions because
rotations will will be in terms of i, j,
00:30:48.200 --> 00:30:56.090
k. And you can think of i, j, i inverse these
kind of things you can think of and i, j,
00:30:56.090 --> 00:31:02.520
i inverse is what is minus j, since the rotation
by 180 degree. So, these kind of things you
00:31:02.520 --> 00:31:14.250
can think of. So, we have obtained that V
4 is also a subgroup of SO 3. So, here r was
00:31:14.250 --> 00:31:20.440
greater than equal to 2, so this is r equals
to 2 case.
00:31:20.440 --> 00:31:25.440
So, if r is greater than equal to 3, then
what happens, so that is where we make
00:31:25.440 --> 00:31:33.330
use of the fact that G is an isometry. G meaning
the elements if the element of my group,
00:31:33.330 --> 00:31:41.930
then it is an isometry that is it keeps
the distance is preserved. So, first thing
00:31:41.930 --> 00:31:48.210
to observe is that if r is greater than equal
to 3, then size of G is twice of r. How do
00:31:48.210 --> 00:31:55.460
we get it? Again from one of the formulas
this one, I obtain that. Just put 1 by 2,
00:31:55.460 --> 00:32:02.620
1 by 2 and 1 by r. So, you will get that
G is of order 2 r.
00:32:02.620 --> 00:32:11.240
And I pick this stabilizer . Now, this stabilizer;
obviously, fixes the pole z stabilizer of
00:32:11.240 --> 00:32:15.780
the pole z of course as to fix the pole z.
And of course, the antipodal pole is minus
00:32:15.780 --> 00:32:25.309
z also has to be fix by this. Therefore, this
is actually a cyclic group of order r; and
00:32:25.309 --> 00:32:39.320
stabilizer can be written as 1, g, g square
and so on . And now this stabilizer this cyclic
00:32:39.320 --> 00:32:48.070
group acts on other poles.
So, I just take one other pole say x. So,
00:32:48.070 --> 00:32:56.790
I consider x, g x, g square x and g to the
power r minus 1 x. I came that that all these
00:32:56.790 --> 00:33:03.320
are distinct; if that were not the case, some
smaller power of g would be fixing x in some
00:33:03.320 --> 00:33:08.580
smaller power which is different from 0. So,
for some i which is different from j, I would
00:33:08.580 --> 00:33:18.930
be getting that g i to the power i i minus
j fixes x, but elements of this all these
00:33:18.930 --> 00:33:26.000
powers are they are allowed to fix only z
and minus z, and certainly axis not z and
00:33:26.000 --> 00:33:32.470
axis not minus z.
And therefore, these are all distinct. And
00:33:32.470 --> 00:33:40.460
now we see the distance preserving observation,
so x and g x, the difference of x and g x
00:33:40.460 --> 00:33:52.270
distance on this sphere the Euclidean distance,
these are all Euclidean distances . So, Euclidean
00:33:52.270 --> 00:34:00.280
distance is same as g x and g square x.
So, I have put f transform this and this both
00:34:00.280 --> 00:34:07.030
by g and so on. So, like this I have. And
I do the same thing with z. The point is g
00:34:07.030 --> 00:34:14.929
to the power i z is z. So, everywhere I am
having z, z, z. So, this kind of picture is
00:34:14.929 --> 00:34:21.151
what I am getting .
So, what is that that means, x, g x, g square
00:34:21.151 --> 00:34:30.900
x g to the power r minus 1 x, they are equidistant
from z. And among themselves also they are
00:34:30.900 --> 00:34:36.560
equidistant. So, picture is something like
there is north pole something like this
00:34:36.560 --> 00:34:51.250
z and each. So, you can imagine z like north
pole, and g to the power i x these are equidistant
00:34:51.250 --> 00:34:59.260
points on equator . You can think of like
this.
00:34:59.260 --> 00:35:17.710
So, like this important clue about how
the poles are configured g, these are coplanar
00:35:17.710 --> 00:35:25.200
as I said they are on the equator, and their
plane is orthogonal to what joins north pole
00:35:25.200 --> 00:35:37.510
with south pole z with minus z. And therefore,
these are forming vertices of regular r-gon.
00:35:37.510 --> 00:35:45.740
And when I observe rotations of this regular
r-gon, I realize that these rotations are
00:35:45.740 --> 00:35:51.290
actually isomorphic to g. This some small
calculation which is involved there, the small
00:35:51.290 --> 00:35:56.810
argument which is involved there. And using
that essentially, what is happening is that
00:35:56.810 --> 00:36:07.160
G is nothing but the regular r-gon with
regular r-gon which is made by which is constitute
00:36:07.160 --> 00:36:17.390
which is x, g x, g square x and so on,
g to the power r minus 1 x these distinct
00:36:17.390 --> 00:36:30.570
points they make regular r-gon and
that is how G becomes isomorphic to D r.
00:36:30.570 --> 00:36:38.820
In other cases as well we have similar kind
of analysis. Let us take this analysis, where
00:36:38.820 --> 00:36:45.010
I had sub case 2 in which case stabilizer
was 2, this stabilizer was 3 size, this was
00:36:45.010 --> 00:36:56.480
size 3. So, again I can use one of the earlier
formulas which one this formula. So, I had
00:36:56.480 --> 00:37:07.160
2, 2, 3.
So, let see 1 plus twice of order of G is
00:37:07.160 --> 00:37:16.730
1 divided by 2 plus 1 divided by 3 plus 1
divided by 4. So, if I do this calculation,
00:37:16.730 --> 00:37:29.980
what do I get here I get 12, and then 6
plus 4 plus 3 which is 13 by 12, and then
00:37:29.980 --> 00:37:47.670
I subtract it. So, 2 divided by this is 13
by 12 minus 1 which is so which gives me
00:37:47.670 --> 00:37:57.610
the so this is 1 divided by 12. So, size of
G is 24. And similarly in the previous case
00:37:57.610 --> 00:38:00.730
size of G is 12.
So, I have actually done the calculation for
00:38:00.730 --> 00:38:07.960
next case. So, in the previous case also the
similar fashion size of group is 12. And because
00:38:07.960 --> 00:38:16.040
of orbits stabilizer formula, we get that
stabilizer of z is having this relation
00:38:16.040 --> 00:38:24.390
orbit of z size is same as 12 divided by size
of stabilizer which is so and this ratio is
00:38:24.390 --> 00:38:28.900
4, because stabilizer z is 3. So, 12 divided
by 3 is 4.
00:38:28.900 --> 00:38:35.190
Then what do we do we pick an elements which
is there in the orbit of z, and elements which
00:38:35.190 --> 00:38:41.340
is different from z and minus z. We can
do that because orbit of z has four elements.
00:38:41.340 --> 00:38:48.630
So, one is z, other is minus z and two more
elements. So, I can pick say w and now I do
00:38:48.630 --> 00:38:56.720
some jugglery with all this. And if this stabilizer
of z is say stabilizer of z is cyclic it has
00:38:56.720 --> 00:39:04.200
three elements. So, let us call it 1, g, g
square. And then I look at the fact of this
00:39:04.200 --> 00:39:13.970
size of stabilizer of z on this element
w on the orbit, and conclude that the three
00:39:13.970 --> 00:39:23.300
distinct elements w, g w and g square w .
And using previous kind of analysis is not
00:39:23.300 --> 00:39:29.930
very difficult to forming to conclude similar
kind of argument using the fact that
00:39:29.930 --> 00:39:37.200
g is distance preserving that z and w, g w,
g square w they actually form a regular tetrahedron.
00:39:37.200 --> 00:39:43.840
So, since they form a regular tetrahedron,
I can have a homomorphism from the finite
00:39:43.840 --> 00:39:56.940
group g that is given to me to the group
of this is group of rotations of tetrahedron
00:39:56.940 --> 00:40:07.160
. I can have that. And again it is not
very difficult for me to conclude that size
00:40:07.160 --> 00:40:14.930
of g is A 4 because you can make count of
elements and all that. So, the group in
00:40:14.930 --> 00:40:25.119
this case is A 4. Again when stabilizer z
is size 4, I can similar kind of analysis
00:40:25.119 --> 00:40:31.520
exactly same kind of analysis, and I can conclude
that G is S 4.
00:40:31.520 --> 00:40:39.380
And similarly when stabilizer is of size
5, I can conclude that G is actually group
00:40:39.380 --> 00:40:45.520
of icosahdedral which is A 5, and then all
the cases are finished. So, I had this case
00:40:45.520 --> 00:40:51.460
of A 5, which is finished . Before that I
had case of cube which is finished and before
00:40:51.460 --> 00:40:54.940
that I had case of tetrahedron, which is also
finished.
00:40:54.940 --> 00:41:06.700
So, we are therefore, back to the original
statement which is that these are all subgroups
00:41:06.700 --> 00:41:16.680
of SO 3. What goes there you carefully examine
what are poles, how stabilizers of other poles
00:41:16.680 --> 00:41:24.550
are acting on those poles and that would eventually
lead either to platonic solids or to a polygons
00:41:24.550 --> 00:41:32.960
which is right there on the equator or a situation
is a Klein's 4-group situation where you have
00:41:32.960 --> 00:41:41.100
the three axis x, y, z axis there are which
are inscribed inside sphere, so that is
00:41:41.100 --> 00:41:48.350
how we prove that SO 3 has finitely many groups.
And those groups are what I have already mentioned
00:41:48.350 --> 00:41:59.270
to you in this lecture. So, this was all about
our group action and how how group action
00:41:59.270 --> 00:42:06.020
is useful for understanding subgroups of SO
3.
00:42:06.020 --> 00:42:14.710
And crucial things has been is the distribution
analysis of distribution of poles one this
00:42:14.710 --> 00:42:21.010
sphere and their orbits under G-action . So,
as you have seen throughout this course
00:42:21.010 --> 00:42:28.990
throughout all these lectures, group action
is quite an important thing. You can use
00:42:28.990 --> 00:42:36.060
group action to understand certain symmetries,
you can use it to understand certain puzzles,
00:42:36.060 --> 00:42:44.700
certain games. And also while doing
more mathematical while exploring other
00:42:44.700 --> 00:42:49.119
aspects of mathematics or the branches of
mathematics such as representation theory,
00:42:49.119 --> 00:42:57.560
matrices and all that, also our groups actions
are imported they are quite inevitable.
00:42:57.560 --> 00:43:06.390
To end this course, I would just mention to
you certain open problems in group theory.
00:43:06.390 --> 00:43:13.700
So, those open problems are most of the
open problems are difficult to state there
00:43:13.700 --> 00:43:19.540
in much more mathematics. Then what has been
covered in these course in these lecture series,
00:43:19.540 --> 00:43:24.859
but nevertheless I manage to find certain
problems which are which can be understood
00:43:24.859 --> 00:43:31.070
in in terms of the parlance in terms of
the terminology that we have used in this
00:43:31.070 --> 00:43:38.500
lecture series, and I would mention it
to you.
00:43:38.500 --> 00:43:45.590
And all these problems I have taken from what
is called Kourovka's notebook. So, Kourovka's
00:43:45.590 --> 00:43:52.430
notebook is maintained by Khukhro and Mazurov,
these are two group theorists. They work
00:43:52.430 --> 00:44:03.171
at this place in Novosibirsk, Russia, Sobolev
Institute of Mathematics. And this Kourovka
00:44:03.171 --> 00:44:08.660
Notebook is currently in its 19th edition
2018, so latest one just came out couple of
00:44:08.660 --> 00:44:13.720
days ago. So, I am just mentioning few problems
which are open problems.
00:44:13.720 --> 00:44:20.110
So, even basic group theory that we just
discussed during these courses has makes
00:44:20.110 --> 00:44:25.500
us capable to understand at least some statements
if not proofs. There are many problems in
00:44:25.500 --> 00:44:30.800
this book for which you need much more background;
and those problems we need lot of probably
00:44:30.800 --> 00:44:37.440
in some cases geometry, lot of algebra, but
these are very straight to understand problems.
00:44:37.440 --> 00:44:44.580
So, I am mentioning these.
So, this is problem 19.11 from Kourovka Notebook
00:44:44.580 --> 00:44:54.350
which says that can you estimate conjugacy
class size for can you estimate number of
00:44:54.350 --> 00:45:02.660
conjugacy classes for finite groups. So, it
says that can we actually express number of
00:45:02.660 --> 00:45:09.820
conjugacy classes as some constant times log
of 2. So, it is quite interesting the count
00:45:09.820 --> 00:45:16.100
to conjugacy classes how many conjugacy classes
are there in terms of log of group size; it
00:45:16.100 --> 00:45:21.500
is interesting.
So, this problem is attributed to E-Bertram
00:45:21.500 --> 00:45:31.130
another problem, problem number 19.30. And
it is in terms of complex characters.
00:45:31.130 --> 00:45:38.870
Remember in one of the earlier lectures, we
have seen what complex characters are. So,
00:45:38.870 --> 00:45:48.200
an element a finite group is set to be vanishing
if chi g is 0. What is chi g, so it is chi
00:45:48.200 --> 00:45:57.440
g is 0 for some irreducible complex character.
So, if on g, if on g, some irreducible character
00:45:57.440 --> 00:46:02.140
vanishes, then you call this element to be
vanishing.
00:46:02.140 --> 00:46:08.200
Now, the question is if you have a finite
group, and then you have another finite simple
00:46:08.200 --> 00:46:16.410
group. And suppose if they have equal orders
and same set of so if the finite symbol group
00:46:16.410 --> 00:46:21.930
and finite group they are same order and the
same set of orders of vanishing elements,
00:46:21.930 --> 00:46:26.500
so you take vanishing elements and the order
of vanishing elements. Suppose, the order
00:46:26.500 --> 00:46:32.770
of vanishing elements these two sets are same,
can you actually conclude that these two groups
00:46:32.770 --> 00:46:42.330
are isomorphic, so it is quite interesting.
Another one which is due to Hooshmand which
00:46:42.330 --> 00:46:51.400
says that we take a finite group of order
n and you factorize this order, so A 1, A
00:46:51.400 --> 00:46:57.480
2, A k, these are integer factorization. So,
it is true that for every factorization. You
00:46:57.480 --> 00:47:05.520
can break, you can find partitions, you can
find subsets A 1, A 2, A k of the group such
00:47:05.520 --> 00:47:13.090
that A 1 has size A 1, A 2 has A 2, A k has
size A k, and every element of G can be obtained
00:47:13.090 --> 00:47:20.131
as some element of A 1 times some element
of A 2 times some element of A k, so again
00:47:20.131 --> 00:47:27.200
it sounds quite an interesting problem.
Last problem which is due to neither than
00:47:27.200 --> 00:47:35.119
but the famous fields medallist the famous
mathematician of our times J.P. Serre. And
00:47:35.119 --> 00:47:45.070
the problem is very simple. It is asks about
infinite simple subgroups of SL 2 Q. So, SL
00:47:45.070 --> 00:47:53.060
is collection of SL 2 is collection of 2 by
2 matrices, and here over rationals. So, 2
00:47:53.060 --> 00:48:01.470
by 2 matrices over rationals, which have determinant
1. In that group, can you find infinite simple
00:48:01.470 --> 00:48:05.510
subgroup; and it is quite surprising that
such problems are still unknown. So, if the
00:48:05.510 --> 00:48:12.200
the straight with the whether this such statements
are true or not is still unknown.
00:48:12.200 --> 00:48:19.730
So, today's I just gave glimpse and each
year this Kourovka's book Kourovka's Notebook,
00:48:19.730 --> 00:48:24.770
they keep adding problems two thousand, so
19th edition these are actually the latest
00:48:24.770 --> 00:48:30.720
editions, the latest edition they have added
some 111 problems; and over all there are
00:48:30.720 --> 00:48:36.280
thousands of problems in Kourovka's book which
are still open. They also keep updating what
00:48:36.280 --> 00:48:42.040
problems have been solved or if some progress
is there what is the status current status
00:48:42.040 --> 00:48:49.410
of that. So, it would be quite interesting
if someone can even comment even small
00:48:49.410 --> 00:48:55.300
comment about these or some examples that
would be quite worth it.
00:48:55.300 --> 00:49:03.140
So, with all this with I am stopping my
lecture series, purpose of all this was to
00:49:03.140 --> 00:49:09.690
make you appreciate what groups are. Although,
in usual courses in many regular group theory
00:49:09.690 --> 00:49:15.610
courses you get to know of various aspects
of group theory, you get to learn what
00:49:15.610 --> 00:49:21.119
Cauchy theorem is, what Sylow subgroups
are. And then are many other aspects how many
00:49:21.119 --> 00:49:26.320
groups how many elements of which order
and all that.
00:49:26.320 --> 00:49:32.790
But we fail to understand connection of
group theory with various real life problems.
00:49:32.790 --> 00:49:38.290
And through this lecture series, I just wanted
to bring that out, so that with the small
00:49:38.290 --> 00:49:44.070
definitions with not much baggage in group
theory we could enjoy, we could appreciate
00:49:44.070 --> 00:49:48.970
why group should be there. And the right from
those basic definitions, we could also come
00:49:48.970 --> 00:49:55.550
we could also see all these hard problems,
all these open problems which are there in
00:49:55.550 --> 00:50:03.060
Kourovka's notebook
There is not much that I would say in this
00:50:03.060 --> 00:50:08.460
course now; I hope you enjoy it all these
lecture series. If you had any query, you
00:50:08.460 --> 00:50:15.020
can write to me, you can ask me through online
inter mode, inter phase, or whenever I
00:50:15.020 --> 00:50:18.300
get live on YouTube, you are always welcome
to ask all the questions.
00:50:18.300 --> 00:50:23.530
Thank you very much, thank you for being for
all these lectures throughout.
00:50:23.530 --> 00:50:24.770
Thank you. .