The question is clear. Can you give me an example of a differentiable curve and a tangent vector on the torus?

And, if its possible, can you give me an example of a differentiable field?

Thank you!

  • $\begingroup$ Do you have a particular expression of the torus in mind? For example, do you think of it as some shape in $\bf{R}^3$? Or as gluing opposite sides of a square? Or as $S^1 \times S^1$? Or as the quotient of $\bf{R} \times \bf{R}$ under the equivalence relation $(x,y) \sim (x+m,y+n)$ for all $(m,n) \in \bf{Z} \times \bf{Z}$? I ask this because expressing a clear answer to your question depends on you giving us a clear idea of how you express the torus. $\endgroup$
    – Lee Mosher
    Nov 30, 2014 at 13:33
  • $\begingroup$ I'm using now the parametrical expresion, which is more useful for my purpose. $R(u, v) = ((R + a \cos (u)) \cos(v),(R + a \cos (u)) \sin (v), a \sin (u))$. But any answer will be thanked :) $\endgroup$
    – MaríaCC
    Nov 30, 2014 at 14:06
  • $\begingroup$ where $a$, $R$ are de radius of the circles of $T^2=S^1xS^1$ $\endgroup$
    – MaríaCC
    Nov 30, 2014 at 14:15
  • $\begingroup$ Did you try anything yourself? $\endgroup$
    – Karl
    Nov 30, 2014 at 14:37

1 Answer 1


Following the comment to use the parametric representation $$R(u,v) = ((R+a \cos(u)) \cos(v), \,\, (R+a \cos(u)) \sin(v), \,\, a \sin(u)) $$ we can think of $R$ as a map with domain $\bf{R}^2$ parameterized by $u,v$.

Any curve $\gamma(t) = (u(t),v(t))$ whatsoever in $\bf{R}^2$, when composed with $R$, gives a curve on the torus. Now pick your favorite curve $\gamma(t)$ and compute. For example, $\gamma(t)=(t,0)$ gives $$R(\gamma(t)) = ( ( R + a \cos(t)) \cos(t), \,\, (R + a \cos(t)) \sin(t), \,\, a \sin(t)) $$

Next, for any point $(u,v) \in \bf{R}^2$ whatsoever, computing the derivative $DR_{(u,v)}$ as a linear function from $\bf{R}^2 \to \bf{R}^3$, and for any vector $\vec V$ in $\bf{R}^2$ whatsoever, the vector $DR_{(u,v)}(\vec V)$ is tangent to the torus at the point $R(u,v)$. Now pick your favorite point $(u,v)$ and vector $\vec V$ and compute. For example, picking $(u,v)=(0,0)$ gives $$DR_{(0,0)} = \pmatrix{0 & 0 & a \\ 0 & R+a & 0} $$ and picking $\vec V = \pmatrix{1 & 0}$ (I'm using row vector notation for some reason) then gives $$DR_{(0,0)} \pmatrix{1 & 0} = \pmatrix{1 & 0} \pmatrix{0 & 0 & a \\ 0 & R+a & a} = \pmatrix{0 & 0 & a} $$ which is tangent to the torus at the point $R(0,0) = (R+a,0,0)$.

Finally, to get a vector field on the torus, first it helps to realize that the map $R(u,v)$ is a quotient map with equivalence relation $(u,v) \sim (u+2 \pi m, v + 2 \pi n)$ (in fact $R$ is a universal covering map from $\bf{R}^2$ to your torus). So, pick your favorite vector field on $\bf{R}^2$ which is invariant under the derivative of this equivalence relation, for example any constant vector field will do, such as $\langle 1,0 \rangle$. Now compute $DR_{(u,v)}(\langle 1,0 \rangle)$ for each $(u,v)$.

  • $\begingroup$ Thank you for your answer. I have a question: How do I compute the derivate $DR_{(u,v)}$. You mean, derivate with respect to $u$ and $v$, then put my favourite point $(u_1,v_1)$ and then put the vector on the final equation I get? (Sorry for my terrible mathematic English.) $\endgroup$
    – MaríaCC
    Nov 30, 2014 at 15:37
  • $\begingroup$ @MariaCC: $DR_{(u,v)}$ is the ordinary matrix of partial derivatives. $\endgroup$
    – Lee Mosher
    Dec 1, 2014 at 3:14

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