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I am new in Alebraic Topology. Given the torus, we say that the fundamental group of the torus is generated by two loops (or more exactly two equivalent classes of loops). One writes $\pi=\mathbb{Z}\times\mathbb{Z}.$

I don't understand, how the null-homotopic loop, which is the constant loop, is generated by the two generators mentioned above. Can somebody provide an explanation? More even so, I don't see how it functions visually, since the two generators are not null-homotopic. More precisely, given a null-homotopic loop on the surface on a base point $x$, how this loop will be generated by the two generators mentioned above?

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    $\begingroup$ The identity is in every subgroup. $\endgroup$ – Angina Seng Dec 28 '18 at 16:06
  • $\begingroup$ Thanks. But how it is generated in the case of a null-homotopic loop ? In the case the loops are traversing the torus around the hole or through the hole, one can say for instance that the product of one loop and its inverse is the constant loop, that is the identity. But I dont see how it works in the case a loop is null-homotopic, since the generators are non null-homotopic. $\endgroup$ – user249018 Dec 28 '18 at 16:14
  • $\begingroup$ What do you think the word "generate" means? $\endgroup$ – Eric Wofsey Dec 28 '18 at 16:20
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Let $\mathbb{T}^2$ denote the torus and choose a basepoint $p \in \mathbb{T}^2$. Then we know that $\pi_1\left(\mathbb{T}^2, p \right) \cong \mathbb{Z} \times \mathbb{Z}$.

Now I think the reason for your confusion is an algebraic one.

Recall that $\mathbb{Z} \times \mathbb{Z}$ has two generators, $a= (1, 0)$ and $b =(0, 1)$. Choose an isomorphism $\psi : \pi_1\left(\mathbb{T}^2, p \right) \to \mathbb{Z} \times \mathbb{Z}$, by surjectivity there exists path classes, $[f], [g] \in \pi_1\left(\mathbb{T}^2, p \right)$ such that $\psi([f]) = a$ and $\psi([g]) =b$. Then since $\psi$ is an isomorphism we have $[f]$ and $[g]$ to be the two generators of $\pi_1\left(\mathbb{T}^2, p \right)$.

Now your question is how the path class of the constant loop $c_p : I \to \mathbb{T}^2$ defined by $c_p(x) = p$ for all $x \in I$, that being $[c_p] \in \pi_1\left(\mathbb{T}^2, p \right)$ is generated by $[f]$ and $[g]$. Well the answer to that is simple: note that $$[c_p] = 1_{\pi_1\left(\mathbb{T}^2, p \right)}$$ that is $[c_p]$ is the identity element of $\pi_1\left(\mathbb{T}^2, p \right)$. Then recall the following definition that we have for exponents in groups.

Definition: In any group $(G, \cdot)$ for any $x \in G$ we define $x^0 = 1_G$ where $1_G$ is the identity element of the group $(G, \cdot)$.

Hence since $[f], [g] \in \pi_1\left(\mathbb{T}^2, p \right)$ and $\pi_1\left(\mathbb{T}^2, p \right)$ is indeed a group, we have $$[f]^0 = [g]^0 = 1_{\pi_1\left(\mathbb{T}^2, p \right)}.$$

Then we have $$\left[c_p\right] = [f]^0 * [g]^0$$ and so the constant path at $p$ is indeed generated by the two generators of $\pi_1\left(\mathbb{T}^2, p \right)$. And since $[c_p]$ is a nullhomotopic loop, since it is a constant loop by definition, the above shows how a product of two non null-homotopic loops yield a null-homotopic loop.


Note that above even though I've gone into quite a bit of detail, the only real fact I'm using is the following algebraic one. If we have a group $G$ and we have $G = \langle A \rangle$ for some subset $A \subseteq G$ then every element $x \in G$ can be written as $x = g_1 \dots g_n \cdot h_1^{-1} \dots h_m^{-1}$ where $g_i, h_i \in G$. In particular if we have $G = \langle c , d \rangle$, that is $G$ is generated by the two elements $c$ and $d$ then we can express $1_G$ as $1_G = c^0 \cdot d^0$.

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Your confusion seems to be about the meaning of the word "generate". By definition, if $G$ is a group and $S\subseteq G$, then the subgroup generated by $S$ is the smallest subgroup that contains $S$. Since a subgroup always contains the identity element, any subset of $G$ (even the empty set!) "generates" the identity element.

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  • $\begingroup$ Thanks. By ''generate'' I meant the group operation on the set of generators. In a group generated by the subset $S$, each element can be written in terms of the generators, which is, elements of $S$. So given the null-homotopic loop on the torus, how can it be put in relation to the elements of $S$, which in our case consists of 2 elements ? $\endgroup$ – user249018 Dec 28 '18 at 16:36
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    $\begingroup$ OK, but the group multiplication is not the only operation in a group! There are two other operations: the identity element and inverses. $\endgroup$ – Eric Wofsey Dec 28 '18 at 16:39
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    $\begingroup$ In particular, one of the operations of a group is an operation which takes no inputs and outputs the identity element. That's how any set "generates" the identity element. $\endgroup$ – Eric Wofsey Dec 28 '18 at 16:40
  • $\begingroup$ Given $S\subset G$, more precisely the group generated by $S$ is defined as $<S>=SS^{-1}$. So you are right about inverses. The thing with the identity element is less obvious. One excepts it very probably by definition...But when it comes to the fundamental group of the torus, are you saying that the null-homotopic loop is generated by the empty set ? Or maybe we can say the following: the multiplication of a geneartor and its inverse gives us the constant loop, which itself is homotopic to null-homotopic loops. Thus we generate the null-homotopic loop from each one of the two generators ? $\endgroup$ – user249018 Dec 28 '18 at 16:55
  • $\begingroup$ I don't know what your notation $SS^{-1}$ is supposed to mean, but it sounds like your definition of "the group generated by $S$" is just wrong (which may not be your fault; you may have been taught a wrong definition!). The correct definition is the one I stated in the answer. An equivalent definition is that the subgroup generated by $S$ is the set of all elements of $G$ that can be obtained by starting with elements of $S$ and repeatedly applying the three operations of the group multiplication, inverses, and the identity element. $\endgroup$ – Eric Wofsey Dec 28 '18 at 17:10
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The subgroup generated by $a$ and $b$ is the set of all elements that can be written as a sequence that consists of nothing but $a$, $b$, and their inverses (e.g $ab$ or $b^{-4}a$). The null sequence is allowed. That is, the empty string (a zero-length sequence) qualifies as "a sequence that consists of nothing but $a$, $b$, and their inverses"; it does not contain anything, so clearly it does not contain anything other than $a$, $b$, and their inverses. In an abelian group with two generators, the group is generated by taking the first generator an integer number of times, and then taking the second generator an integer number of times. And zero is an integer. Given two non null-homotopic loops $a$ and $b$, the constant loop is generated by taking $a$ zero times, then taking $b$ zero times. If you think of a group in terms of group actions, the identity is generated by not doing anything. Doing nothing at all is, at least as far as mathematicians are concerned, an action. Or, in the words of Geddy Lee, if you choose not to decide, you still have made a choice.

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