I'm reading Differential Forms in Algebraic Topology by Bott and Tu and I'm a little confused about the notation $H^{*-n}(M)$. I don't see a definition of this listed anywhere in the text, though I can sort of infer from context what's meant by the notation.

  • I know that $H^k(M)$ is the $k$th cohomology class on $M$ (equivalence class of $k$-forms).
  • I know too that $H^*(M)$ is the direct sum of all cohomology classes (i.e. $\bigoplus_{i=0}^\infty H^i(M)$ ).
  • The first place in the text I can recall seeing this notation (or similar notation) is when they talk about the Poincar${\acute {e}}$ Lemma where they state an isomorphism $H_c^{*+1}(\mathbb{R}^n \times \mathbb{R}^1) \simeq H_c^*(\mathbb{R}^n)$. My guess here is that it has to do with how the dimensions of the forms get mapped to cohomology classes of different spaces (i.e. in this example stated, since $\mathbb{R}^n \times \mathbb{R}^1$ is essentially the same as $\mathbb{R}^{n+1}$, the isomorphism is somehow stating that $H_c^k(\mathbb{R}^n \times \mathbb{R}^1) \simeq H_c^{k-1}(\mathbb{R}^n)$?
  • This notation is also used too in the context of vector bundles. If $\pi:E\to M$ is a rank-$n$ vector bundle on a manifold $M$ of dimension $k$, then Bott and Tu have stated that $H_{cv}^*(E) \simeq H^{*-n}(M)$ where the $cv$-distinction is for forms of compact vertical support. Is this to indicate that a compact-vertically supported $l$-form gets naturally sent to an $l-n$ form on $M$ via the projection $\pi_*$?

I may have just answered my own question, but I'm self-studying and it would be nice to have verification.

  • 6
    $\begingroup$ I think it means something like $H^i(E) \cong H^{i-n}(M)$ for all $i$ where this makes sense. $\endgroup$
    – user113102
    Commented Sep 17, 2019 at 20:27

1 Answer 1


This community wiki solution is intended to clear the question from the unanswered queue.

In fact is is not unusual to write $H^*(M) = \bigoplus_{i=0}^\infty H^i(M)$. Then $$H^{*-n}(M) = \bigoplus_{i=0}^\infty G_i$$ where $G_i = 0$ for $i = 0,\ldots,n-1$ and $G_i = H^{i-n}(M)$ for $i \ge n$. Roughly speaking, this gives a new grading to $H^*(M)$.

As user113102 comments, it is simpler to regard $*$ as a variable. You could also write $i, m$ or something else instead of $*$.

  • $\begingroup$ Thank you for the answer. I suppose it is more straightforward to say something of the form "$H^k(E) \simeq H^{k-n}(M)$ for all $k=n, n+1, \ldots$"? $\endgroup$
    – Mnifldz
    Commented Sep 18, 2019 at 18:44
  • $\begingroup$ That is correct! $\endgroup$
    – Paul Frost
    Commented Sep 18, 2019 at 22:51

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