# What is $H^{*-n}(M)$?

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.

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

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 $$*$$.
• 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 $k=n, n+1, \ldots$"? Commented Sep 18, 2019 at 18:44