The kunneth formula gives that $H^k(X \times Y) = \bigoplus_{i+j = k} H^{i}(X) \otimes H^{j}(Y)$, where $X$ and $Y$ are both manifolds. I wonder whether this is true on the cochain level. More specifically we have $\pi_1$, $\pi_2$ projections onto $X$ and $Y$, is it true that every $k$ form on $X\times Y$ is of the form $\sum\pi_1^{\ast}(w_1) \land \pi_2^{\ast}(w_2)$?


No, this is false. The easiest case to see it is $k=0$: then you would be claiming that every smooth function $f$ on $X\times Y$ can be written as a finite sum of functions $(g_i\circ\pi_1)\cdot (h_i\circ\pi_2)$ for smooth functions $g_i$ and $h_i$ on $X$ and $Y$ respectively. But this is false; for instance, writing $f_x(y)=f(x,y)$, this implies that the functions $f_x$ span a finite-dimensional vector space (since they are all linear combinations of the $h_i$), which is not true for all $f$ (exercise: find an explicit smooth function $f$ on $\mathbb{R}\times\mathbb{R}$ such that infinitely many of the functions $f_x$ are linearly independent).

  • $\begingroup$ But in Bott-Tu's algebraic topology page 34, they claim that every form on $\mathbb R^n \times \mathbb R$ is a uniquely a linear combination of $\pi^{\ast} \phi f$ and $\pi^{\ast} \phi fdt$ where $\pi$ is the projection onto $\mathbb R^n$ and $w$ is a form on $\mathbb R^n$, $f$ a function on $\mathbb R^n \times \mathbb R$ $\endgroup$ – Keith May 25 at 4:38
  • $\begingroup$ The statement you asked about originally includes no such function $f$. Note that crucially $f$ is a function on $\mathbb{R}^n\times\mathbb{R}$, rather than a function on just one of the coordinates. $\endgroup$ – Eric Wofsey May 25 at 4:45
  • $\begingroup$ Would you please explain that why it would be true with such a function? $\endgroup$ – Keith May 25 at 4:52
  • $\begingroup$ A form on $\mathbb{R}^{n+1}$ is just a sum of smooth functions times expressions of the form $dx_{i_1}dx_{i_2}\dots dx_{i_k}$. Each of those terms is either of the form $\pi^*\phi f$ or $\pi^*\phi f dt$, depending on whether $dt$ is one of the $dx_{i_j}$. $\endgroup$ – Eric Wofsey May 25 at 5:16
  • $\begingroup$ It's not clear what Bott and Tu mean when they claim this decomposition is "unique", and it is not actually unique in most obvious senses. But you don't actually need uniqueness: you just need $K$ to be well-defined and linear, and you can achieve that by always using the decomposition given by the coordinate forms as in my previous comment. $\endgroup$ – Eric Wofsey May 25 at 5:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.