Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $D = 0$?.

share|improve this question
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
add comment

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 $\sqrt f|_{z_0}=\sqrt0=0$.

share|improve this answer
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
add comment

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, $D = 0$.

share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.