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I've a question about the following. Take an affine space $(E,V)$ and consider the tangent space $T_aE$ at a point $a$.

From John Lee book "Introduction to smooth manifolds" chapter 3 there exist a canonical isomorphism between the "geometric tangent space" $\mathbb R_a^n$ at point $a$ and the vector space of derivations at $a$, namely:

$$D_v|_a f = D_vf(a) = \left . \frac {d} {dt} \right |_{t=0} f(a + tv)$$

Now the vector $v \in \mathbb R_a^n$ can be understood as an element of translation vector space $V$. Therefore I believe there is a canonical/natural isomorphism between $V$ and $T_pE$.

Carlo C
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  • $V$ does not contain infinitesimal vectors. The translations equivalent to elements of $T_pE$ are infinitesimal. To get an element of $V$ from $T_pE$, one must integrate along a path in $E$, parametrized by $p$ with integrand a map $p \rightarrow T_pE$. (Equivalently, specify a tangent bundle on $E$ and integrate the associated differential equation(s) to get flowlines. Note that there is no reason to assume the flowlines correspond to a constant translation everywhere in the space -- constancy may require changing to coordinates that normalize the flowlines.) – Eric Towers Mar 26 '24 at 16:01
  • @EricTowers, one has a natural map $V\to T_pE$ by $v\mapsto d/dt\vert_{t=0}(p+tv)$. Because $V$ acts freely and transitively, this map is an isomorphism. The translation action means that given $p\in E$ there is a natural identification of $E$ with $V$ which relates the constant vector fields on $V$ with those which generate the translation action of $V$ on $E$. No need to change coordinates. – J.V.Gaiter Mar 26 '24 at 17:11
  • Is it possibile to show that $v \mapsto d/dt\vert_{t=0}(p+tv)$ is isomorphism without introducing any chart for the affine space ? – Carlo C Mar 26 '24 at 18:10
  • Which smooth atlas do you use on $E$? – Paul Frost Mar 27 '24 at 00:18
  • @PaulFrost We can use an atals on $E$ given by a single affine chart (i.e. picked a point in $E$ and a basis for the translation vector space $V$, then a bijective linear map is defined that send each $V$ basis vector into a different n-tuple in $\mathbb R^n$ ) – Carlo C Mar 27 '24 at 07:33

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Let me begin by a remark concerning the geometric tangent space $\mathbb R^n_a$ of $\mathbb R^n$ at $a$. Lee defines it as $$\mathbb R^n_a = \{ a \} \times \mathbb R^n $$ with the vector space structure canonically induced from $\mathbb R^n$, i.e. the unique vector space structure making the canonical bijection $\phi_a : \mathbb R^n \to \mathbb R^n_a, \phi_a(v) = (a,v)$, an isomorphism. Alternatively, we can say $\{a\}$ is a vector space (this is trivial, each single point set is a vector space in a unique way) and regard $\mathbb R^n_a$ as the product vector space of $\{a\}$ and $\mathbb R^n$.

Actually $\mathbb R^n_a$ is an affine subspace of $\mathbb R^n \times \mathbb R^n$ and we use $\phi_a$ to identify it with the vector space $\mathbb R^n$.

One could also define $\mathbb R^n_a = \mathbb R^n$, and many authors do so. It is a matter of taste! A benefit of Lee's definition is that $\mathbb R^n_a$ contains the information to which $a \in \mathbb R^n$ it belongs. The $\mathbb R^n_a$ are in fact pairwise disjoint.

Lee associates to each $(a, v) \in \mathbb R^n_a$ the map $D_v \mid_a : C^\infty(\mathbb R^n) \to \mathbb R, D_v \mid_a(f) = \frac {d} {dt} |_{t=0} f(a + tv)$ which is a derivation at $a$ and proves that $(a, v) \mapsto D_v \mid_a$ is vector space isomorphism $\mathbb R^n_a \to T_a \mathbb R^n$.

Changing perspective, one could also say that for each $a \in \mathbb R^n$ one gets a vector space isomorphism $\iota_a : \mathbb R^n \to T_a \mathbb R^n$ via $v \mapsto D_v \mid_a$. The only difference to Lee's approach is that we have to specify separately the point $a$, whereas Lee trivially knows $a$ for the elements of $\mathbb R^n_a$.

Let us come to your affine space $(E,V)$. For each $p \in E$ there is a bijection $\beta_p : V \to E, \beta_p(v) = p + v$. A smooth structure on $E$ is generated by a single chart of the form $\psi \circ \beta_p^{-1} : E \to \mathbb R^n$, where $\psi : V \to \mathbb R^n$ is an isomorphism. It is the unique smoth structure making $\psi \circ \beta_p^{-1}$ a diffeomorphism. It is easy to see that resulting smooth structure does not depend on the choices of $p$ and $\psi$.

Note that also $V$ has a canonical smooth structure making $\psi$ a diffeomorphism; again it does not depend on the choice of $\psi$.

What should be the "geometric tangent space" $E_p$ of $E$ at $p \in E$? I think one should take $$E_p = \{p \} \times V$$ which is identfied with $V$ via $\phi_p : V \to E_p, \phi_p(v) = (p,v)$. Alternatively one can take $$E_p = V$$ as above for $\mathbb R^n$. It seems that you prefer the second variant and want to get a canonical isomorphism $V = E_p \to T_pE$. Anyway it does not make much difference to work with $V$ or with $\{p \} \times V$ here.

Let $\lambda_0 : V \to T_0V$ and $\kappa_p : V \to T_pE$ be the unique isomorphisms making the following diagram commutative:

$\require{AMScd}$ \begin{CD} V @>{\kappa_p}>> T_pE \\ @V{id_V}VV @VV{d(\beta_p^{-1})_p}V \\ V @>{\lambda_0}>> T_0V \\ @V{\psi}VV @VV{d\psi_0}V \\ \mathbb R^n @>>{\iota_0}> T_0 \mathbb R^n \end{CD}

It is easy to see that $\lambda_0$ does not depend on the choice of $\psi$, thus it is a canonical isomorphism.

Also $d(\beta_p)_0 : T_0V \to T_pE$ may be regarded as the canonical isomorphism, simply because the smooth structure on $E$ was defined via $\beta_p$.

This shows that $\kappa_p$ has every right to be regarded as the canonical isomorphism.

Paul Frost
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    Thanks for your answer. I believe each $\mathbb R_a^n$ cannot be an affine subspace of $\mathbb R^n$ since they have the same dimension. Each $\mathbb R_a^n$ is just a copy of $\mathbb R^n$ attached to the point $a$. Any of them is canonically identified with $\mathbb R^n$ as vector space by mean of $\phi_a$. – Carlo C Mar 27 '24 at 17:24
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    About the tangent space at a point $p$ of affine space $(E,V)$, I believe we can take either $E_p = {p } \times V$ or $T_pE$ itself (the latter is the set of derivations at $p$ that has got a vector space structure). – Carlo C Mar 27 '24 at 17:36
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    @CarloC My remark concerning $\mathbb R^n_a$ was nonsense. I corrected it. The tangent space of $E$ at $p$ is of course $T_pE$. But as for $\mathbb R^n$ (which can also be regarded as an affine space $(E,V) = (\mathbb R^n,\mathbb R^n)$) we can define a variant of the "geometric tangent space" and this is ${p} \times V$, or alternatively $V$. – Paul Frost Mar 28 '24 at 06:48
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    Ok, so even for $\mathbb R^n$ as affine space $E = (\mathbb R^n, \mathbb R^n)$ the tangent space at $p$ is $T_pE$ (i.e. the set of derivations at $p$ that is a vector space on its own). In this specific case the variant of $T_pE$ (as "geometric tangent space at $p$") is actually ${p} \times V$ (or alternatively $V$). – Carlo C Mar 28 '24 at 09:34
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    @CarloC Yes, the tangent space of each manifold $M$ should be understood as $T_pM$ , the vector space of derivations at $p$. There are alternative constructions (e.g. equivalence classes of curves through $p$). For submanifolds of $\mathbb R^N$, in particular for $\mathbb R^N$ itself, there is a "geometric" variant which one may denote as Euclidean tangent space. See https://math.stackexchange.com/q/2982185 and https://math.stackexchange.com/q/4421786. This "geometric variant" can also be used (with the obvious modificatons) for submanifolds of vector spaces $V$ or affine spaces $(E,V)$. – Paul Frost Mar 28 '24 at 09:50
  • From links you referenced, the name you used for the Euclidean tangent space at point $p$ is $\tilde T_pM$ which is different name from $T_pM$ (the latter is the standard notation/name for the vector space of derivations at point $p$). In the case of $E=(\mathbb R^n, \mathbb R^n)$, $\tilde T_pE$ is actually another "geometric" variant of the ${p} \times \mathbb R^n$ geometric tangent space at $p$. – Carlo C Mar 28 '24 at 10:45
  • @CarloC What I denoted by $\tilde T_pM$ for a submanifold $M \subset \mathbb R^N$ is a linear subspace of $\mathbb R^N$. It is a "geometric" variant of the tangent space and it is canonicall isomorphic to $T_pM$, but it is available only for submanifolds of Euclidean spaces. For open subsets $M \subset \mathbb R^N$ we get $\tilde T_pM = \mathbb R^N$. This corresponds to my remark that many authors take $\mathbb R^n_a = \mathbb R^n$ instead of ${a} \times \mathbb R^n$ . – Paul Frost Mar 28 '24 at 10:50
  • @CarloC I think I have said all I can say. – Paul Frost Mar 28 '24 at 10:52
  • Take the set of curves $u(t)$ on the submanifold $M \subset \mathbb R^n$ such that $u(0)=p$. Then the set of derivatives $u′(t)$ evaluated at $t=0$ form a linear subspace of $\mathbb R^n$ as vector space. Such a vector space is $\tilde T_pM$, right? – Carlo C Mar 29 '24 at 13:46
  • @CarloC Yes, this is correct! – Paul Frost Mar 29 '24 at 18:25