---
title: 'lecture 01: singular simplices and chains'
description: standard simplices, singular simplices, chains, the boundary operator, and the definition of singular homology
date: '2026-10-08'
modified: 2026-10-09 09:01:29 GMT-04:00
tags:
  - course
  - mit
  - math/topology
created: '2026-10-08'
published: '2026-10-08'
pageLayout: default
slug: courses/18.905-fall-2016/lectures/01
permalink: https://aarnphm.xyz/courses/18.905-fall-2016/lectures/01.md
generator:
  quartz: v4.6.0
  hostedProvider: Cloudflare
  baseUrl: aarnphm.xyz
full: https://aarnphm.xyz/llms-full.txt
---
Miller opens with the syllabus and then builds singular homology from scratch in four definitions. The source is [[courses/18.905-fall-2016/resources/mit18_905f16_lec1|lecture 1]] of [[courses/18.905-fall-2016/index|18.905]].

## spaces we care about

Euclidean space $\mathbb{R}^n$, the sphere $S^n = \{x \in \mathbb{R}^{n+1} : \lvert x \rvert = 1\}$, real projective space $\mathbb{RP}^n = S^n / (x \sim -x)$, the Stiefel manifold $V_k(\mathbb{R}^n)$ of orthonormal $k$-frames (topologized inside $(S^{n-1})^k$), and the Grassmannian $\mathrm{Gr}_k(\mathbb{R}^n)$ of $k$-planes, a quotient of the Stiefel manifold by taking spans. $\mathrm{Gr}_1(\mathbb{R}^n) = \mathbb{RP}^{n-1}$.

All of these are manifolds. The spheres, projective spaces, Stiefel manifolds, and Grassmannians above are compact; $\mathbb{R}^n$ is noncompact for $n > 0$. The course ends with Poincaré duality for closed manifolds.

The method: probe a space $X$ by maps into it from simple spaces. Maps from simplices give singular homology. Maps from spheres give homotopy groups (18.906).

## the four definitions

**Standard simplex.** For $n \geq 0$,

$$
\Delta^n = \Big\{ \sum t_i e_i : \sum t_i = 1,\ t_i \geq 0 \Big\} \subseteq \mathbb{R}^{n+1},
$$

the convex hull of the standard basis $e_0, \dots, e_n$. The $t_i$ are barycentric coordinates. For $n \geq 1$, the face inclusion $d^i : \Delta^{n-1} \to \Delta^n$ ($0 \leq i \leq n$) is the affine map sending vertices to vertices in order and skipping $e_i$.

**Singular simplex.** A singular $n$-simplex in $X$ is a continuous map $\sigma : \Delta^n \to X$. $\mathrm{Sin}_n(X)$ is the set of all of them. For $n \geq 1$, precomposition with $d^i$ gives the $i$-th face $d_i \sigma = \sigma \circ d^i \in \mathrm{Sin}_{n-1}(X)$.

**Chains.** $S_n(X) = \mathbb{Z}\,\mathrm{Sin}_n(X)$ is the free abelian group on singular $n$-simplices. An $n$-chain is a finite sum $\sum a_i \sigma_i$ with $a_i \in \mathbb{Z}$. For $n < 0$, $S_n(X) = 0$. The point of passing to the free group: for a loop $\sigma$, “the boundary vanishes” should read $d_0 \sigma - d_1 \sigma = 0$, and that difference is not a simplex.

**Boundary and homology.** In degree $n \geq 1$, the boundary operator is

$$
d\sigma = \sum_{i=0}^{n} (-1)^i d_i \sigma,
$$

extended additively to $d : S_n(X) \to S_{n-1}(X)$. In degree zero, set $d : S_0(X) \to S_{-1}(X) = 0$ to be the zero map. A point has no boundary in this chain complex. Cycles $Z_n(X) = \ker d$, boundaries $B_n(X) = \operatorname{im}(d : S_{n+1} \to S_n)$, and

$$
H_n(X) = Z_n(X) / B_n(X).
$$

The quotient makes sense because of theorem 1.5: $d^2 = 0$, every boundary is a cycle. The proof is problem set 1.

A chain complex is a graded abelian group $\{A_n\}$ with maps $d : A_n \to A_{n-1}$ satisfying $d^2 = 0$. Cycles, boundaries, and homology are defined for any chain complex the same way.

## what to expect from the groups

$Z_n(X)$ and $B_n(X)$ are free abelian groups, as subgroups of the free abelian group $S_n(X)$. Their ranks depend on the space and the degree. For a one-point space, there is exactly one singular simplex in each nonnegative degree, so $S_n(*) \cong \mathbb{Z}$ and both subgroups have rank at most one. This corrects the lecture’s blanket claim that cycles and boundaries are uncountably generated.

The quotient $H_n(X)$ need not be free, though it is finitely generated for the spaces above. For the torus $T$, $H_1(T) \cong \mathbb{Z} \oplus \mathbb{Z}$, generated by the two coordinate loops.

The sphere:

$$
H_q(S^n) =
\begin{cases}
\mathbb{Z} & q = n > 0 \\
\mathbb{Z} & q = 0,\ n > 0 \\
\mathbb{Z} \oplus \mathbb{Z} & q = n = 0 \\
0 & \text{otherwise.}
\end{cases}
$$

The $n = 0$ row is the only one that looks odd: $S^0$ is two points, so $H_0$ counts two components.

## problems

1. Prove $d^2 = 0$ on $S_n(X)$. The input is the simplicial identity $d_i d_j = d_{j-1} d_i$ for $i < j$, which comes from $d^j d^i = d^i d^{j-1}$ on standard simplices. Pair the terms of $\sum_{i,j} (-1)^{i+j} d_i d_j$ across $i < j$ and $i \geq j$.
2. Check that a constant map $\Delta^1 \to X$ is a cycle, and that it is a boundary.

