1,625 reputation
1728
bio website none
location United Kingdom
age 23
visits member for 2 years, 9 months
seen Jul 22 at 14:39

Currently studying for an Msc in Mathematics


Jun
22
comment Axiom of Limitation of Size Reference Request
Thanks, I managed to track down a reference to von Neumann's collected works, I imagine it would be far to convenient if there happened to be a translation somewhere
May
23
comment Showing that $\mathbb{R}\times [0,1]$ is quasi-isometric to $\mathbb{R}$
@DanielFischer Just the standard metric (euclidean) I think as it is not specified.
May
23
comment Showing that $\mathbb{R}\times [0,1]$ is quasi-isometric to $\mathbb{R}$
@drhab I have edited in the definition of quasi-isometric, does this help?
May
22
comment Do we sometimes have to go “each way” separately for iff proofs?
I find this question interesting, although it may have a simple answer, I could imagine there being some crazy counterexample- I'm thinking of some sort of independence proof?
May
22
comment Showing that triangles in $\mathbb{Z}$ are thin
@DerekHolt Ah yes of course, thanks for the help! :) If you wish to post as an answer I would gladly accept
May
22
comment Showing that triangles in $\mathbb{Z}$ are thin
@DerekHolt thanks. Could you explain your first line a bit please?
May
19
comment Showing geodesics in $\mathbb{H}^2$ have unique common perpendicular or common endpoint
OK, but how do I go about doing this?
May
19
comment Showing geodesics in $\mathbb{H}^2$ have unique common perpendicular or common endpoint
Is there no way I can do this with intermediate value theorem like you suggested? Something like taking a $\gamma_3$ that intersects $\gamma_2$ perpendicularly and then vary the endpoints of $\gamma_3$ somehow to get the intersection with $\gamma_1$ to be perpendicular?
May
19
comment Showing geodesics in $\mathbb{H}^2$ have unique common perpendicular or common endpoint
I'm not sure sorry, my basic algebra is pretty bad! How do I calculate that?
May
19
comment Showing geodesics in $\mathbb{H}^2$ have unique common perpendicular or common endpoint
Sorry I is quite hard to explain what I mean without pictures!
May
19
comment Showing geodesics in $\mathbb{H}^2$ have unique common perpendicular or common endpoint
My geodesics are circles that intersect the boundary perpendicularly so I was being stupid taking straight line not through the center as it does not intersect perpendicularly sorry. So do I take $\gamma_3$ which intersects $\gamma_2$ perpendicularly. Then vary the other endpoint and by IVT there will be a point which intersects $\gamma_1$ perpendicularly.
May
19
comment Showing geodesics in $\mathbb{H}^2$ have unique common perpendicular or common endpoint
Ok so I say that $\gamma_1$ is the horizontal line through the origin and take $\gamma_2$ such that they don't share endpoints on $S_\infty$. Now I need to find geodesic intersecting them both perpendicularly. So any vertical line will intersect $\gamma_1$ perpendicularly can I then show that it intersects $\gamma_2$ perpendicularly at some point?
May
19
comment Showing geodesics in $\mathbb{H}^2$ have unique common perpendicular or common endpoint
No I am using the interior of the unit disc in $\mathbb{R}^2$ sorry I should have mentioned that I will edit my question
May
16
comment Baumslag–Solitar $B(1,2)$ is not hyperbolic
@N.Owad No, could you give me reference/ explain to me?
May
5
comment Proving that the tensor product is generated by $a\otimes b$
@Magdiragdag yes sorry that is what I meant, the stuff generated by the tensors $a\otimes b$. I think there is a problem with this approach though. Say we let $K$ be the module generated by stuff of the form $a\otimes b$. Then take $x,y\in A\otimes B-K$. Then I see no reason as to why we couldn't have $x+y\in K$ and $x+y\neq 0$ which would be a problem for the map.
Apr
30
comment $\mathbb{C}\otimes_\mathbb{C} \mathbb{C} \cong \mathbb{R}\otimes _\mathbb{R} \mathbb{C}$
Is there a slight subtly here in that we can write a map $f:\mathbb{C}\otimes \mathbb{C}\rightarrow \mathbb{C}$ such $f(a\otimes b)=ab$ and then show that this map is injectvive/surjective. However do we also need to check that this map is well defined, which is a problem as we don't know what $a\otimes b$ is and so we have to define $g:\mathbb{C}\times \mathbb{C}\rightarrow \mathbb{C}$ such that $g(a,b)=a\otimes b$ and then use the universal property to show that the map from before actually exists and is well defined?
Apr
30
comment Showing that $\mathbb{Z}/2\mathbb{Z}[\mathbb{Z}/2\mathbb{Z}]$ is semi-simple
@JackSchmidt Are you saying that $\{0,g\}$ is an ideal? I am now confused, don't we have $\{0,g\}g=\{0,e\}$ hence this is not an ideal?
Apr
29
comment Showing that $\mathbb{Z}/2\mathbb{Z}[\mathbb{Z}/2\mathbb{Z}]$ is semi-simple
Thanks! Sorry for the pretty trivial question
Apr
29
comment Showing that $\mathbb{Z}/2\mathbb{Z}[\mathbb{Z}/2\mathbb{Z}]$ is semi-simple
So $\{0,e\}$ is not an ideal as $\{0,e\}g\not\subset \{0,e\}$ but $\{0,g\}$ is an ideal right as is $\{0,e+g\}$ so does this choice work?
Apr
29
comment Showing that $\mathbb{Z}/2\mathbb{Z}[\mathbb{Z}/2\mathbb{Z}]$ is semi-simple
oh yeah, oops. Yeah if I just leave the last summand off I think I am ok right?