I am trying to find an example of a diagonal linear operator T in L(H) H is hilbert space that is bounded but not compact and also one which is compact but not Hilbert-Schmidt. any Ideas??
Where diagonal means Ten=§en where § is the eigenvalue and en is on orthonormal basis.
Ok thanks I get that as I thought.
I am trying to show that the fundamental group of a product of Topological Spaces is isomorphic to the product of fundamental groups:
pi1(X x Y , (p,q)) -> pi1(X,p) x pi1(Y,q)
I can understand this but want to construct the isomorphism and am...
I am doing some revision and trying to do fundamental groups and I was wondering if the fundamental group of the following space is {1} i.e. all loops based p are homotopic.
fundamental group of (X,p) = D^2\{(x,0) : 0<=x<=1} where p=(-1,0)
I am struggling to properly understand the concept of a well-founded set.
Is this well founded, d = {{x},{x,y},{x,y,z}}
because there exists an element of d i.e. {x} = e
such that d n e = 0 ?
Just because it is R^3 with a slice missing along the z axis and 1 unit up in y-axis however witth more reflection I think that it would have Z as fudamental group because as before the space is still path connected since a path to any to point can go around the missing slice but loops going...
I am not sure since a loop from x to x in
R^3\{(x; y; z) | x = 0; y = 0; 0 <= z <= 1}
is a similar situation to
R^3\{(x; y; z) | x = 0; y = 0; z = 0}
As they can always avoid the unit line on z axis that is missing as it were.
Also is R^3\{(x; y; z) | x = 0; 0 <= y <= 1}
isomorphic to...
for part (ii)
if [x]=[f(x)] in RPn
then by the definition
λx=f(x) so that they are in the same equivalence class therefore since f is linear x is an eigenvector and λ is the eigenvalue
Yes that is very helpful thank you, in my mind I was trying to calculate it from one point. So for example you say that if X is path connected and x not equal to y in X, then the fundamental group at x is the same as at y.
So for example R^3\(0,0,0)?
This I believe is path connected however...
Also R^3 with the x-axis plane removed, can that be homotopic equivalent to a torus and therefore have fundamental group Z disjoit union Z as with a torus.
I have done some more research and now I understand that the fundamental group of X is the group consisting of the homotopic equivalence classes of loops of base x in X. But this must be expressed in algebraic terms, now I understand that {2} is not a group but I beleiev there are two different...
if you had to give a reason for this would you say that is because it is a 3 dimensional manifold.
Also does this mean that if X2 = {(x; y; z) ∈ R^3 | x not equal to 0}
then pi1(X2) = {2} since it splits R^3 into two sections