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I want to solve the following exercise.

Let A be an $\mathbb{R}$-algebra. A $derivation$ on A is a $\mathbb{R}$-linear map $D: A \to A$ obeying the Leibniz rule $$ D(ab) = D(a)b + aD(b) $$ for all $a,b \in A$.

Now let M be a manifold. Show that the algebra $$ C(M) = \{ f: M \to \mathbb{R} : f \textrm{ continuous } \} $$ has no non-trivial derivations, i.e. for every linear map $D : C(M) \to C(M)$ obeying the Leibniz rule, it follows that $D = 0$.

Hint: Use that every $f \ge 0$ can be written as a square.

I have some difficulties in understanding. I am currently reading about manifolds and tangent spaces, and I read that the tangent space could be defined as the space of all point derivations $D : C^{\infty}(M) \to \mathbb{R}$, but this space is more than just the trivial derivation $D = 0$, does it make a huge difference if I am considering smooth vs. just continuous functions? And in the exercise, how does the hint help me? I can write every $f \ge 0$ as $(\sqrt{f})^2$, and then $$ D(g^2) = 2gD(g) = gD(g+g) $$ with $g = \sqrt{f}$, does it follow that $D = 0$?.

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Hint: For fixed $x \in M$, you can assume that $f(x) = 0$ because if $f_1 = f - f(x)$, then $df_1 = df$. Now use that $df(x) = 2 g(x) dg(x)$. –  Michael Joyce Aug 17 '12 at 15:33
    
@MichaelJoyce That trick seems to fail, because if $f\geq 0$, it is not true that $f_1\geq 0$. I think it can be fixed, though. –  Thomas Andrews Aug 17 '12 at 15:39
    
@ThomasAndrews: The fix should be to write $f_1 = h_1 - h_2$ with $h_i$ both continuous, both $\geq 0$, and such that $h_i(x) = 0$ for both. Then argue as above that $dh_1(x) = dh_2(x) = 0$, so $df(x)=0$. This is essentially the same argument that allowed us to assume $f \geq 0$ in the first place. –  Michael Joyce Aug 17 '12 at 16:14
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2 Answers 2

First $D(const)=0$, so we can assume $f(z_0)=0$. (or you can use $D(f)=D(f-f(z_0))$).

Then $D(f)|_{z_0}=2D(\sqrt f)|_{z_0}(\sqrt f)|_{z_0}=0$ for $\sqrt f|_{z_0}=\sqrt0=0$.

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but you just looked at a single point $z_0$, how do you infere that $Df = 0$ for every point? –  Stefan Aug 17 '12 at 15:59
    
let $z_0$ move over the manifold and it's ok. Actually, this computation suits for $f<0$. Because we can extend $D$ to $C^\infty(M)\otimes_{\mathbb{R}}\mathbb{C}=C^\infty(M;\mathbb{C})$ and get a derivation again. Such a derivation can act on the complex root $\sqrt{f}$ of $f$, even if $f$ is negative at some point. –  user18537 Aug 17 '12 at 16:08
    
Oh, a tiny mistake $C(M)\otimes\mathbb{C}$ and $C(M;\mathbb{C})$, here we consider the continuous functions. –  user18537 Aug 17 '12 at 16:18
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It is clear from the definition that $D(1) = D(1.1) = 2 D(1)$, hence $D(1) = 0$. Furthermore, in a similar manner, we have $D f^2 (x) = 2 f(x) D f (x)$. From this we have $D |f| (x) = 2 \sqrt{|f(x)|} D \sqrt{|f|} (x)$

Choose $c\in \mathbb{R}$, and let $f_{c,+} (x) = \max (f(x), c)$, $f_{c,-} (x) = \min (f(x), c)$. For any $x, c$ we have $f(x) = f_{c,+} (x) + f_{c,-} (x)$, and both $f_{c,+}, f_{c,-}$ are continuous. Clearly $f_{c,+}(x) \geq c$, $f_{c,-}(x) \leq c$ for all $x$.

By linearity, we have $D f = D f_{c,+} + D f_{c,-}$. Furthermore, we have $D f_{c,+}(x) = D (f_{c,+}-c)(x) = (2 \sqrt{f_{c,+}(x)-c} ) D (\sqrt{f_{c,+}-c}) (x)$. It follows that $D f_{f(x_0),+}(x_0) = 0$, for any $x_0$. Similarly (using $-D f_{c,-}(x) = D (c-f_{c,-})(x)$), we obtain $D f_{f(x_0),-}(x_0) = 0$. Consequently we have $D f (x_0) = D f_{f(x_0),+}(x_0) + D f_{f(x_0),-}(x_0) = 0$. Since $x_0$ was arbitrary, we have $Df = 0$. Since $f$ was arbitrary, $D = 0$.

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