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2

I will give an answer without using the fact that the fixed point set of an isometry is a totally geodesic submanifold. Instead we use that isometry preserves Levi-Civite connection. Hence we have (for any $t\in I$) $$\tag{1} df_{\gamma(t)}( \nabla _{\gamma'(t)} \gamma'(t) )= \nabla _{df_{\gamma(t) } \ \gamma'(t)}df_{\gamma(t) } \gamma'(t) = \nabla _{\gamma'(...


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It turns out there are already two answers to this question on MSE: All connected closed subgroups in $SO(3)$ and Connected lie subgroup of $SO(3)$


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Here is a simple example. Take the cylinder with boundary $ C = S^1 \times [0,1] $ Then leave the two boundary circles identical (isometric) but slightly distort the interior of the cylinder (say by adding a bump somewhere) so that the new cylinder with boundary $ C' $ has no isometries. Then glue infinitely many copies of $ C' $ end to end. The resulting ...


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If $v_0$ is a fixed point of $h(v) = Av+w$ then $h(v)$ can be represented as $h(v) = A(v-v_0) + Av_0 + w = A(v-v_0) + v_0$ where $A$ is a rotation matrix. But expression $A(v-v_0)+v_0$ is a rotation by matrix $A$ around $v_0$ because it represents translation of $v_0$ to the origin, rotation of the translated $v$ (which is $v-v_0$) by $A$, then translating ...


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Having a more explicit description of horocycles helps here. In the disk model, horocycles are exactly the circles on the Riemann sphere contained in $\mathbb{D} = \{ z \mid |z| < 1\}$ and which are tangent to the circle $S = \{ z \mid |z| = 1\}$. The isometries of $\mathbb{D}$ in this model are the Möbius transformations that send $\mathbb{D}$ and $S$ to ...


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In complex notation, as you noticed, $H_t(z)=\alpha_t z+\beta_t$. By definition, the center of rotation is the unique solution $c_t$ of $H_t(c_t)=c_t$. That is: $$c_t=\frac{\beta_t}{1-\alpha_t}$$ And the instantaneous center $I$, if it exists, is represented by complex number $$c=\lim_{t\rightarrow 0}\frac{\beta_t}{1-\alpha_t}\tag{1}$$ First, notice that the ...


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At the risk of being a bit confusing, I will use $\otimes$ to denote the Kronecker product. We have $$ \varphi(X) = \pmatrix{X \otimes e_1 & X \otimes e_2 & \cdots & X \otimes e_n}. $$ From there, we find that $$ \langle \varphi(X), \varphi(Y) \rangle = \sum_{i=1}^n \langle X \otimes e_i, Y \otimes e_i\rangle = \sum_{i=1}^n \langle X,Y \rangle = ...


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HINT: The solution of @MathematicsStudent1122 is quite salvageable, just consider the space $$\mathbb{Z} \cup (\mathbb{Z}+ \frac{1}{3})\cup (\mathbb{Z}+\frac{1}{2})$$


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Formally, in a quotient $G/H$, two elements $g,g'$ are in the same coset iff (def.) $$ab^{-1} \in H $$. In our case, per this definition, for $G:=SL(2, R)$ ; $H:=\{ I,-I\}$,: Two matrices $M, M' $are equivalent, aka , in the same coset iff : $$MM'= \pm Id$$ , i.e., if $MM'=Id$ or $MM'= -Id $ Multiplying each equation by $M^{-1}$, this comes down to $M, M'$ ...


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Your notation is somewhat confusing. Correct me if I am wrong, but I think with $\ell^n_1$ and $\ell^n_\infty$ you mean the space of sequences of length $n$ equipped with the $1$-norm and the max-norm, respectively? In other words, you consider the space $\mathbb{R}^n$ (assuming the sequences consist of real numbers) equipped either with norm $\|x\|_1=\sum_{...


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