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HINT: The solution of @MathematicsStudent1122 is quite salvageable, just consider the space $$\mathbb{Z} \cup (\mathbb{Z}+ \frac{1}{3})\cup (\mathbb{Z}+\frac{1}{2})$$
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 $\|x\|_1=\sum_{... 2 I will give an answer without using the fact that the fixed point set of an isometry is a totally geodesic submanifold. Instead we use that isometry preserves Levi-Civite connection. Hence we have (for any$t\in I\$) $$\tag{1} df_{\gamma(t)}( \nabla _{\gamma'(t)} \gamma'(t) )= \nabla _{df_{\gamma(t) } \ \gamma'(t)}df_{\gamma(t) } \gamma'(t) = \nabla _{\gamma'(... 2 Having a more explicit description of horocycles helps here. In the disk model, horocycles are exactly the circles on the Riemann sphere contained in \mathbb{D} = \{ z \mid |z| < 1\} and which are tangent to the circle S = \{ z \mid |z| = 1\}. The isometries of \mathbb{D} in this model are the Möbius transformations that send \mathbb{D} and S to ... 1 It turns out there are already two answers to this question on MSE: All connected closed subgroups in SO(3) and Connected lie subgroup of SO(3) 1 If v_0 is a fixed point of h(v) = Av+w then h(v) can be represented as h(v) = A(v-v_0) + Av_0 + w = A(v-v_0) + v_0 where A is a rotation matrix. But expression A(v-v_0)+v_0 is a rotation by matrix A around v_0 because it represents translation of v_0 to the origin, rotation of the translated v (which is v-v_0) by A, then translating ... 1 In complex notation, as you noticed, H_t(z)=\alpha_t z+\beta_t. By definition, the center of rotation is the unique solution c_t of H_t(c_t)=c_t. That is:$$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}$$First, notice that the ... 1 At the risk of being a bit confusing, I will use \otimes to denote the Kronecker product. We have$$ \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 = ...