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Let $X$ be a topological space (assume additional assumptions if needed) and denote by $\mathcal O _X$ its sheaf of $\Bbbk$-valued continuous functions where $\Bbbk$ is $\mathbb{R}$ or $\mathbb{C}$ with standard topology.

Then, as it is done in the differentiable setting or in algebraic geometry, one can define the following objects $$T_X:=\mathscr{Der}_\Bbbk (\mathcal O_X,\mathcal O_X)$$ the tangent sheaf, i.e. the sheaf of $\Bbbk$-linear derivations of $\mathcal O_X$ with values in $\mathcal O_X$ (on local sections, $\Bbbk$-linear maps $D:\mathcal O_X(U)\to\mathcal O_X(U)$ satisfying Leibniz: $D(f\cdot g)=f\cdot Dg + g\cdot Df$), and $$\Omega_X^1:=\mathcal I/\mathcal I^2$$

the sheaf of differentials, where $\mathcal I$ is the ideal sheaf of $X$ embedded diagonally $\Delta:X\hookrightarrow X\times X$ into $X\times X$ (i.e. $\mathcal I(U)=$ functions in $\mathcal O_{X\times X}(U)$ that are zero on every point of $\Delta(X)\subset X\times X$).

Well, what can be said about these two sheaves? Anything interesting at all?

Also, is there any relationship between $T_X$ and the "tangent microbundle" $\tau_X$ in case $X$ is a topological manifold?

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    $\begingroup$ $\mathcal I/\mathcal I^2$ is often zero. In particular, it is zero for $X=\mathbb R$. $\endgroup$
    – André Henriques
    Commented Apr 10, 2019 at 20:31
  • $\begingroup$ Oh that's true, I didn't think about that $\endgroup$
    – Qfwfq
    Commented Apr 10, 2019 at 20:39
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    $\begingroup$ $T_X$ is always $0$. If $D$ is a derivation and $f$ is a function, then for every point $x$ $Df$ vanishes at $x$; it suffices to prove this when $f(x)=0$, and in that case $f=gh$ where both $g(x)=0=h(x)$, so $Df=gDh+hDg$ vanishes at $x$. $\endgroup$
    – Tom Goodwillie
    Commented Apr 10, 2019 at 23:49
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    $\begingroup$ @TomGoodwillie May I request that you post this as an "answer" so that the OP may accept it? $\endgroup$
    – Theo Johnson-Freyd
    Commented Apr 11, 2019 at 3:40
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    $\begingroup$ Just a comment, in the smooth situation, there is a third sheaf we could consider, namely the sheaf of Kähler differentials (as in algebraic geometry). This sheaf does not coincide with the sheaf of (correct) differential forms, but the dual of that sheaf is $T_X$ (with $\mathcal{O}_X$ the sheaf of smooth functions) and the dual of $T_X$ is the sheaf of (correct) differential forms. $\endgroup$
    – Ingo Blechschmidt
    Commented Apr 17, 2019 at 13:58

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Your $𝑇𝑋$ is always $0$. If $𝐷$ is a derivation and $𝑓$ is a function, then for every point $𝑥$ $𝐷𝑓$ vanishes at $𝑥$; it suffices to prove this when $𝑓(𝑥)=0$, and in that case $𝑓=𝑔ℎ$ where $𝑔(𝑥)=0=ℎ(𝑥)$, so $𝐷𝑓=𝑔𝐷ℎ+ℎ𝐷𝑔$ vanishes at $𝑥$.

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  • $\begingroup$ why does it suffice to prove this for $f=0$? $\endgroup$
    – user2520938
    Commented Apr 12, 2019 at 6:52
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    $\begingroup$ @user2520938 because any function $g$ can be decomposed as $g=g(x) + (g-g(x))$. The first summand is constant hence vanishes when applying a derivation, the second summand is a function with value zero at $x$. $\endgroup$
    – Michael Bächtold
    Commented Apr 12, 2019 at 8:41
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    $\begingroup$ @user2520938 $D$(const.)=0 holds always. It follows from the Leibniz rule of derivations and $\mathbb{k}$-linearity. $\endgroup$
    – Michael Bächtold
    Commented Apr 12, 2019 at 10:48

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