# What do the differential k-forms on a product manifold look like?

I am interested in how I could express $\Omega^k( M \times N)$ in terms of $\Omega^i(M)$ and $\Omega^j(N)$ for $i,j = 0,1, \ldots k$. Is there a nice relation?

This question arose in the context of the Freund-Rubin solution to the bosonic equations of motion for 11-dimensional supergravity, where one posits a space-time geometry $AdS_4 \times S^7$ and a 4-form field strength proportional to the volume form on $AdS_4$, considered as an element of $\Omega^4(AdS_4 \times S^7)$. In verifying that such a geometry/field strenth do indeed satisfy the equations of motion, it is necessary to consider its Hodge dual in $\Omega^7(AdS_4 \times S^7)$. I suspect that this is simply the volume form on $S^7$, but can't show this.

I also suspect that the answer to my more general question concerning $\Omega^k( M \times N)$ is not necessary to determine the special case in the second paragraph, but I am quite curious.

Denote by $p_M: M \times N \longrightarrow M$ and $p_N: M \times N \longrightarrow N$ the canonical projections. Then you get an induced bilinear map from $\Omega^i(M) \times \Omega^j(N) \longrightarrow \Omega^{i+j}(M \times N)$ by taking the $\wedge$-product of the pullbacks with $p_M$ and $p_N$. Hence you have a linear map $$\Omega^i(M) \otimes \Omega^j(M) \longrightarrow \Omega^{i+j}(M \times N)$$ which turns out to be injective. Taking now all $i$ and $j$ with $k = i+j$ gives you already a big part of the differential forms but not all of them (take a look at $k = 0$ where this is not yet surjective). Nevertheless, the image of this map is dense in the natural Frechét topology of differential forms. This gives a bijection of $\Omega^k (M \times N)$ with the sum of the completed projective tensor product of the individual differential forms of the correct degrees.