1,662 reputation
1730
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location United Kingdom
age 23
visits member for 2 years, 11 months
seen 2 days ago

Currently studying for an Msc in Mathematics


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
revised Showing geodesics in $\mathbb{H}^2$ have unique common perpendicular or common endpoint
added 73 characters in body
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
19
asked Showing geodesics in $\mathbb{H}^2$ have unique common perpendicular or common endpoint
May
18
accepted $\mathbb{C}\otimes_\mathbb{C} \mathbb{C} \cong \mathbb{R}\otimes _\mathbb{R} \mathbb{C}$
May
18
accepted Question on Frobenious Reciprocity
May
17
accepted Baumslag–Solitar $B(1,2)$ is not hyperbolic
May
16
comment Baumslag–Solitar $B(1,2)$ is not hyperbolic
@N.Owad No, could you give me reference/ explain to me?
May
16
asked Baumslag–Solitar $B(1,2)$ is not hyperbolic
May
9
awarded  Popular Question
May
8
revised $\mathbb{1}\uparrow_H^{G}$ is the permutation representation on $G/H$
added 23 characters in body
May
8
asked $\mathbb{1}\uparrow_H^{G}$ is the permutation representation on $G/H$
May
7
revised Showing that $g$ and $g^{-1}$ are conjugate iff $\chi(g)$ is real
deleted 2 characters in body
May
7
asked Showing that $g$ and $g^{-1}$ are conjugate iff $\chi(g)$ is real
May
5
asked Question on Frobenious Reciprocity
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
accepted Proving The Diamond Lemma