Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

So I'm trying to use Van Kampen theorem to prove that a space is null-homotopic.

The thing is I got it down to this $\langle a\mid a=1\rangle$, however I'm confused what does this mean.

For calculating the the torus you get it down to this $\langle a,b\mid a^{-1}b^{-1}ab=1\rangle \cong \mathbb{Z} \times \mathbb{Z}$.

But, was thinking does the brackets mean $\langle 1\rangle \cong \mathbb{Z}$ . I'm confused as normally the $\langle$ , $\rangle$ brackets means the generating set.

share|cite|improve this question
$\langle a\;|\;a=1 \rangle$ is the group with one generator $a$, which satisfies $a = 1$, hence the trivial group. BTW, what does it mean for a space to be null-homotopic? Did you mean simply connected? – Alexander Thumm Mar 14 '12 at 10:09
In the brackets we have generators G and relations R: <G|R>. If there is no relation and one generator, then this group is isomorphic to $\mathbf{Z}$. – HbCwiRoJDp Mar 14 '12 at 10:10

More in general with Van Kampen theorem one have to deal with notation of the following kind

$$\langle g_1,\dots,g_s \mid m_1(g_1,\dots,g_s) = n_1(g_1,\dots,g_s);\dots; m_k(g_1,\dots,g_s) = n_k(g_1,\dots,g_s) \rangle$$ where $m_i(g_1,\dots,g_s)$ and $n_j(g_1,\dots,g_s)$ are strings containing as symbols $g_i$ or $g_i^{-1}$.

This notation is called a finite presentation of a group: it simply denote the biggest group having $s$ generators, named $g_1,\dots,g_s$ for which the relations

$$m_i(g_1,\dots,g_s) = n_i(g_1,\dots,g_s)$$ hold. More formally the notation above denote the quotient group of the free group with $s$-generators by the smaller normal subgroup containing the elements of the form

$$m_i(g_1,\dots, g_s) \left(n_i(g_1,\dots,g_s)\right)^{-1}\ \text{.}$$

In this group the equalities $m_i=n_i$ hold and every other group for which $m_i=n_i$ is a quotient of this group, thus the attribute bigger.

Some additional notes: in your examples $\langle a \mid a = 1 \rangle$ means the group with one generator with is equal to the identity, the group with this property is the trivial group $\{1\}$ that contain just the identity (so this proves that you space is indeed simply connected. In the case of the torus $\langle a,b \mid aba^{-1}b^{-1} = 1\rangle$ is also the group $\langle a,b \mid ab=ba\rangle$, this group is the biggest group having two generators which commute, i.e. it is the free abelian group with two generators.

share|cite|improve this answer

The notation $\langle a|a=1\rangle$ would usually mean the group generated by $a$ subject to the relation $a=1$, so in fact it is the trivial group and your space is null-homotopic as you require. What the notation is really saying is that it's the quotient of the free group on $a$ by itself, which is clearly trivial.

share|cite|improve this answer
Here I'm reading "null-homotopic" as "simply connected", i.e. every loop is null-homotopic. Thanks to Alexander Thumm for pointing this out in his comment. – Matthew Pressland Mar 14 '12 at 10:13

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.