---
title: Product manifolds
summary: A product of manifolds is a manifold whose tangent space stacks its factors', and whose retraction acts factor by factor; derivatives in its charts are block matrices of the factors'.
depends_on: [retraction, derivatives-in-charts]
---

# Product manifolds

Several quantities taken together, such as a pose and a velocity, or every state in a window of an
estimator, are one point of a product manifold. This note shows that the product inherits a
retraction from its factors, and how derivatives in its charts break into blocks.

## The product and its tangent space

Let $\manifold_1, \dots, \manifold_k$ be manifolds of dimensions $\dimension_1, \dots, \dimension_k$, each with a
[retraction](/math/retraction.md). Their product $\manifold = \manifold_1 \times \cdots \times \manifold_k$
is a manifold of dimension $\dimension = \sum_i \dimension_i$, whose points are tuples $\point = (\point_1, \dots, \point_k)$. Its
tangent space at $\point$ is the direct sum of the factors', identified with $\R^\dimension$ by stacking:

$$
	\T_\point \manifold = \T_{\point_1}\manifold_1 \oplus \cdots \oplus \T_{\point_k}\manifold_k \;\cong\; \R^\dimension,
	\qquad \tangent = (\tangent_1; \dots; \tangent_k) . @tag(stack)
$$

(The $\oplus$ of vector spaces here is the direct sum, not the retraction.)

> **Proposition (The product retraction).** If each $R^i$ is a retraction on $\manifold_i$, then
>
> $$
> 	R_\point(\tangent) = \bigl(R^1_{\point_1}(\tangent_1), \dots, R^k_{\point_k}(\tangent_k)\bigr)
> $$
>
> is a retraction on $\manifold = \manifold_1 \times \cdots \times \manifold_k$, its inverse is
> $\otherPoint \ominus \point = (\otherPoint_1 \ominus \point_1; \dots; \otherPoint_k \ominus \point_k)$, and it is flat when every factor is.

<Proof>

$R_\point(0) = (\point_1, \dots, \point_k) = \point$. The derivative of $R_\point$ at $0$ is block diagonal, its $i$-th
block the derivative of $R^i_{\point_i}$ at $0$, which is $I_{\dimension_i}$; so it is $I_\dimension$. The inverse and
flatness follow factor by factor.

</Proof>

## Derivatives in product charts

Let $\map : \manifold \to \otherManifold$ be smooth, with $\otherPoint = \map(\point)$. Because a step in the product chart
moves each factor in its own chart, the [derivative in charts](/math/derivatives-in-charts.md) of $\map$ is
a row of blocks,

$$
	\D \map(\point) = \bigl[\, \D_1 \map(\point) \;\; \cdots \;\; \D_k \map(\point) \,\bigr],
	\qquad
	\D_i \map(\point)\, \tangent_i = \frac{\dd}{\dd t}\Big|_{t=0} \bigl(\map(\point_1, \dots, \point_i \oplus t \tangent_i, \dots, \point_k) \ominus \otherPoint\bigr),
	@tag(blocks)
$$

each the derivative in the $i$-th factor's chart with the other factors held fixed. A map into a
product is a column of blocks in the same way, one per factor of the output. For a scalar $f : \manifold \to \R$, the
chart gradient stacks the factors' partial gradients, $g = (g_1; \dots; g_k)$, and the chart
Hessian is the block matrix whose $(i, j)$ block differentiates $g_i$ along the $j$-th factor:
$H_{ij}(\point)\, \tangent_j = \frac{\dd}{\dd t} g_i(\point_1, \dots, \point_j \oplus t \tangent_j, \dots, \point_k)$ at
$t = 0$.

In particular, the retraction of the product is, as a map of the pair $(\point, \tangent)$, the factors'
retractions side by side, and its derivatives of every order are block diagonal in the factors.
Nothing about a product's derivatives requires more than its factors' derivatives and the
bookkeeping of [](eq:stack).