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melanie science forum beginner
Joined: 15 Jul 2005
Posts: 13

Posted: Thu Jul 13, 2006 1:38 am Post subject:
Question from an old algebraic topology qualifying exam



Hello,
I am struggling on an old qual, it seems difficult.
I am not sure how to solve these.
Any help would be appreciated.
Problem;
Let Y be a subset of 3 dimentinal Euclid space such that
Y = {(x,y,z) in R^3 : 9 =< x^2 + y^2 + z^2 =< 16,
and x^2 + z^2 >= 1, y^2 + z^2 >= 1 }
Let the boundary of Y be X and consider the continuous
function
f: X > X ; (x, y, z) > (y, x, z)
1)Find the integer coefficient homology group H_* (X; Z).
2)Suppose f induces selfhomomorphism on the first
dimentional integer coefficient homology group
f_* : H_1 (X; Z) > H_1 (X;Z).
Find the characterisic polynomial of its matrix. 

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Lee Rudolph science forum Guru
Joined: 28 Apr 2005
Posts: 566

Posted: Thu Jul 13, 2006 11:07 am Post subject:
Re: Question from an old algebraic topology qualifying exam



melanie <melanie@yahoo.com> writes:
Quote:  Let Y be a subset of 3 dimentinal Euclid space such that
Y = {(x,y,z) in R^3 : 9 =< x^2 + y^2 + z^2 =< 16,
and x^2 + z^2 >= 1, y^2 + z^2 >= 1 }
Let the boundary of Y be X

Okay. So can you describe X in some other way, as an
example of a space you are (or should be) familiar with?
One (natural) way to get this description is to look
at the description above and take it bit by bit. Thus:
{(x,y,z) in R^3 : 9 =< x^2 + y^2 + z^2 =< 16} describes
the closed region between two spheres (of radii 3 and 4),
so *its* boundary is precisely those two spheres.
The inequalities x^2 + z^2 < 1, y^2 + z^2 < 1
(complementary to the other two inequalities in the
description of Y) each describe an open region that
is a tubular (product) neighborhood of a straight
line in R^3, namely, the y axis and the x axis
respectively. Y is the region between those two
spheres with those open regions both *removed*,
and (because the radius of the tubes is 1 which
is less than 3) its boundary X is the union of
the following parts: (1) the two spheres, each
with four (round) disks removed; (2) four cylindrical
surfaces, each bounded by one (round) circle on the
outer sphere and one on the inner sphere. So X is
a surface (2dimensional manifold). *What* surface?
Count and find out.
Quote:  and consider the continuous
function
f: X > X ; (x, y, z) > (y, x, z)

This function, being the restriction to X of an orthogonal
linear transformation A of R^3, and the spheres, cylinders,
etc., mentioned above all behaving nicely with respect to
A, can now be describedEASILYin terms of how it
permutes the two 4punctured 2spheres in (1) and the
four cylinders in (2)
..
Quote:  1)Find the integer coefficient homology group H_* (X; Z).

As soon as you have answered "*What* surface?", you will know
this *abstractly*. If you think a bit more, you should know it
*concretely*, in the sense that you will actually be able to
find, first, a geometrically described set of 1cycles on X whose
homology classes generated the homology group, and, second,
the (fairly obvious!) relations among themso you'll have the
group given by generators and relations.
Quote:  2)Suppose f induces selfhomomorphism on the first
dimentional integer coefficient homology group
f_* : H_1 (X; Z) > H_1 (X;Z).

Why "suppose"? It does, and that's an end on't.
Quote:  Find the characterisic polynomial of its matrix.

Continue as indicated.
Lee Rudolph 

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melanie science forum beginner
Joined: 15 Jul 2005
Posts: 13

Posted: Thu Jul 13, 2006 4:22 pm Post subject:
Re: Question from an old algebraic topology qualifying exam



Quote:  melanie <melanie@yahoo.com> writes:
Let Y be a subset of 3 dimentinal Euclid space such
that
Y = {(x,y,z) in R^3 : 9 =< x^2 + y^2 + z^2 =< 16,
and x^2 + z^2 >= 1, y^2 + z^2 >= 1 }
Let the boundary of Y be X
Okay. So can you describe X in some other way, as an
example of a space you are (or should be) familiar
with?
One (natural) way to get this description is to look
at the description above and take it bit by bit.
Thus:
{(x,y,z) in R^3 : 9 =< x^2 + y^2 + z^2 =< 16}
describes
the closed region between two spheres (of radii 3 and
4),
so *its* boundary is precisely those two spheres.
The inequalities x^2 + z^2 < 1, y^2 + z^2 < 1
(complementary to the other two inequalities in the
description of Y) each describe an open region that
is a tubular (product) neighborhood of a straight
line in R^3, namely, the y axis and the x axis
respectively. Y is the region between those two
spheres with those open regions both *removed*,
and (because the radius of the tubes is 1 which
is less than 3) its boundary X is the union of
the following parts: (1) the two spheres, each
with four (round) disks removed; (2) four cylindrical
surfaces, each bounded by one (round) circle on the
outer sphere and one on the inner sphere. So X is
a surface (2dimensional manifold). *What* surface?
Count and find out.

I thimk it is the torus.
Quote: 
and consider the continuous
function
f: X > X ; (x, y, z) > (y, x, z)
This function, being the restriction to X of an
orthogonal
linear transformation A of R^3, and the spheres,
cylinders,
etc., mentioned above all behaving nicely with
respect to
A, can now be describedEASILYin terms of how it
permutes the two 4punctured 2spheres in (1) and the
four cylinders in (2)
.
1)Find the integer coefficient homology group H_*
(X; Z).
As soon as you have answered "*What* surface?", you
will know
this *abstractly*. If you think a bit more, you
should know it
*concretely*, in the sense that you will actually be
able to
find, first, a geometrically described set of
1cycles on X whose
homology classes generated the homology group, and,
second,
the (fairly obvious!) relations among themso you'll
have the
group given by generators and relations.

If X is torus then,
H_0 (X; Z) = Z
H_1 (X; Z) = Z (+) Z
H_q (X; Z) = 0 ( q > 1 )
Quote:  2)Suppose f induces selfhomomorphism on the first
dimentional integer coefficient homology group
f_* : H_1 (X; Z) > H_1 (X;Z).
Why "suppose"? It does, and that's an end on't.
Find the characterisic polynomial of its matrix.
Continue as indicated.

I trnaslated wrong.
Quote:  Find characrteristic equation of its matrix is right.

Is this just a characteristic equation of next matrix?
[0, 1, 0]
[1, 0, 0]
[0, 0, 1]
i.e. its characteristic equation is
(lamda)^2*(1lamda) + 1 = 0


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Lee Rudolph science forum Guru
Joined: 28 Apr 2005
Posts: 566

Posted: Thu Jul 13, 2006 6:51 pm Post subject:
Re: Question from an old algebraic topology qualifying exam



melanie <melanie@yahoo.com> writes:
....
Quote:  Y is the region between those two
spheres with those open regions both *removed*,
and (because the radius of the tubes is 1 which
is less than 3) its boundary X is the union of
the following parts: (1) the two spheres, each
with four (round) disks removed; (2) four cylindrical
surfaces, each bounded by one (round) circle on the
outer sphere and one on the inner sphere. So X is
a surface (2dimensional manifold). *What* surface?
Count and find out.
I thimk it is the torus.

I think you should draw a picture.
Lee Rudolph 

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melanie science forum beginner
Joined: 15 Jul 2005
Posts: 13

Posted: Fri Jul 14, 2006 5:04 pm Post subject:
Re: Question from an old algebraic topology qualifying exam



Quote:  melanie <melanie@yahoo.com> writes:
...
Y is the region between those two
spheres with those open regions both *removed*,
and (because the radius of the tubes is 1 which
is less than 3) its boundary X is the union of
the following parts: (1) the two spheres, each
with four (round) disks removed; (2) four
cylindrical
surfaces, each bounded by one (round) circle on
the
outer sphere and one on the inner sphere. So X is
a surface (2dimensional manifold). *What*
surface?
Count and find out.
I thimk it is the torus.
I think you should draw a picture.
Lee Rudolph

I drew a picture, it looks like two spherical shell
with four round holes. Does this look like a
familiar surface?
I have a hard time to figure out.
Could you explain in details?
Thanks; 

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Lee Rudolph science forum Guru
Joined: 28 Apr 2005
Posts: 566

Posted: Sat Jul 15, 2006 7:43 pm Post subject:
Re: Question from an old algebraic topology qualifying exam



melanie <melanie@yahoo.com> quoted a description of a surface,
which I rephrased thus:
Quote:  Y is the region between those two
spheres with those open regions both *removed*,
and (because the radius of the tubes is 1 which
is less than 3) its boundary X is the union of
the following parts: (1) the two spheres, each
with four (round) disks removed; (2) four
cylindrical
surfaces, each bounded by one (round) circle on
the
outer sphere and one on the inner sphere. So X is
a surface (2dimensional manifold). *What*
surface?

Later I recommended that melanie draw a picture. Now:
Quote:  I drew a picture, it looks like two spherical shell
with four round holes

in each of the two.
No, it doesn't: it looks like that PLUS the lateral
surfaces of four right circular cylinders; each of
these four pieces of cylinder has one boundary
circle on the outer sphere (where it concides
with the boundary of one of the round holes on
that sphere) and one on the inner sphere (where
it coincides with the boundary of one of the round
holes on *that* sphere).
Quote:  Does this look like a
familiar surface?

Do you know the "classification of surfaces"? If you
do, then you have enough information to know which
surface it is (and it is familiar to you); if not,
then it may not be familiar to you. Even if it
is unfamiliar, you do have enough information to
(in theory) compute the homological information you
originally asked about, but you may not have enough
sophistication (in practice).
Lee Rudolph 

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