# Question on how differential form as defined for subsets of $\mathbb R^n$ and integration on them in Spivak, Calculus on Manifolds

If $V$ is a vector space, denote by $\Lambda^k(V)$ the space of alternating multilinear maps from $V^k$ to $\mathbb R$, i.e. the space of alternating $k$-tensors. Also for a point $p \in \mathbb R^n$ denote by $\mathbb R^n_p$ the tangent space $\{ p \} \times \mathbb R^n$ at $p$.

In Spivak, Calculus on Manifolds, he defines a k-form, or simply differential form, to be a map $\omega$ such that for every $p \in \mathbb R^n$ we have $\omega \in \Lambda^k(\mathbb R^n_p)$. Then for a differentiable (where by differentiable he always means $C^{\infty}$) $f : \mathbb R^n \to \mathbb R^m$ and the linear transformation $Df(p) : \mathbb R^n \to \mathbb R^m$ he defines $f_{\ast} : \mathbb R^n_p \to \mathbb R^m_{f(p)}$ by $$f_{\ast}(v_p) = (Df(p)(v))_{f(p)}$$ Then he defines the pull-back of $k$-forms on $\mathbb R^m$ by $$f^{\ast}\omega(p)(v_1, \ldots, v_k) = \omega(f(p))(f_{\ast}(v_1), \ldots, f_{\ast}(v_k))$$ for $p \in \mathbb R^n, v_1,\ldots, v_k \in \mathbb R^n_p$; thus giving differential forms on $\mathbb R^n$.

What is missing here is that he is not defining differential forms for subsets of $\mathbb R^n$ or for functions $f : A \to \mathbb R^m$ with $A \subseteq \mathbb R^n$. The latter could be fixed easily, as $f$ is assumed to be differentiable it has a tangent space everywhere on $A$. But I am not sure how to define differential forms on arbitrary subsets of $\mathbb R^n$, if we could say simply that every subset has $\mathbb R^n_p$ at every point $p$ as its tangent space? For open sets, this might be clear, but not if they might have a boundary or look even wilder.

Maybe these observations are fixed if we restrict to subsets which are differentiable manifolds, but this is to much of a machinery, as up to this point manifolds are not introduced. So there should be a simpler solution?

Later in the chapter on the Fundamental theorem he writes

If $\Omega$ is a $k$-form on $[0,1]^k$, then $\omega = fdx^1 \wedge \cdots \wedge dx^k$ for a unique function $f$.

and uses this to define the integral of $\omega$ over $[0,1]^k$.

But how could $\omega$ be a differential form on $[0,1]^k$? It was just defined for $\mathbb R^k$. So does this means that $\omega(p) \in \Lambda_k(\mathbb R^k_p)$ for $p \in [0,1]^k$? But how is this justified, looking up a more general definition (like on wikipedia) a $k$-form for some subset is a section in the $k$-th exterior power of the cotangential space; but that means that for each point we have a multilinear map on tangent vectors. But is $\mathbb R^n_p$ the tangent space of $[0,1]^k$. For open sets, I see that we could have tangent vectors in all directions, but this cube is closed, hence has a boundary, and there I do not see that the tangent space is also isomorphic to $\mathbb R^k$?

Then later he proceeds:

If $\omega$ is a $k$-form on $A$ (guess $A \subseteq \mathbb R^n$, but this is not mentioned) and $c$ is a singular $k$-cube in $A$ we define $$\int_c \omega = \int_{[0,1]^k} c^{\ast}\omega.$$

With the above definitions we must have for $c^{\ast}\omega$ with $p \in [0,1]^k$ $$\omega(c(p))(c_{\ast}(v_p))$$ where $c_{\ast}(v_p) = (Dc(p)\cdot v)_{c(p)}$ with $Dc(p) : \mathbb R^k \to \mathbb R^n$. But again, this has the same drawback as written above, what happens on the boundary. The map $c$ is differentiable, so it has a tangent space at its boundary by definition. But as it is continuous $c([0,1]^k))$ is a closed set with boundary, so again it is not clear how the tangent space should be defined at the boundary?

So there seems something missing? Does he subtly assumes that for every $A \subseteq R^n$ the tangent space is $R^n_p$ for each $p \in \mathbb R^n$?

Spivak does everything inside $\mathbb{R}^n$. At that point in the book, the author hasn't even defined manifolds and submanifolds, so I don't see why one would worry about subsets $A \subset \mathbb{R}^n$ that have boundaries or look particularly crazy: the tangent space at $a \in A \subset \mathbb{R}^n$ is the only thing it could be, namely $\mathbb{R}^n_a$. The author did define what it means for a function $f:A \to \mathbb{R}^m$ to be differentiable (extend to an open set containing $A$ and differentiate), so it seems natural to assume that when he speaks of a differentiable $k$-form on $A$ he means an assignment of an element of $\Lambda^k(\mathbb{R}^n_a)$ to each $a \in A$ such that the local expression of the form is given by differentiable functions $A \to \mathbb{R}$.
• Thanks for your answer. Guess by local expression you mean the extension to an open subset, but then this is not a function $A \to \mathbb R$, it is either a function $U \to \Lambda^k(R_a^n)$ or a function $U \times (R_a^n)^k \to \mathbb R$, where $A \subseteq U$ is open, or not? And which one of the alternatives you mean? – StefanH Mar 21 '16 at 19:05
• @Stefan I mean the functions $f_{i_1,\ldots,i_k}:A \to \mathbb{R}$ given by $\omega = \sum_{i_1 < \cdots < i_k} f_{i_1,\ldots,i_k} dx^{i_1}\wedge \cdots \wedge dx^{i_k}$. (At each point $a \in A$, $f_{i_1,\ldots,i_k}(a)$ is the $(i_1,\ldots,i_k)$-th coordinate of $\omega(a)$ in the usual basis of $\Lambda^k(\mathbb{R}^n_a)$.) When I wrote the answer I had the case $k = n$ in mind, in which only one such function appears. – Alex Provost Mar 21 '16 at 19:15
• Okay, makes sense. By the way, first your first sentences, that for arbitrary $A \subseteq \mathbb R^n$ the tangent space is simply $\mathbb R^n_a$ already answers my question; but your remarks about differentiability are interesting anyway! – StefanH Mar 21 '16 at 19:20