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Normal-class calculation for real projective space

Example

Assume AC. In $H^{*} ({R P}^{9}; F_{2}) = F_{2} [a] / (a^{10})$ , the normal total class is $\overset{ˉ}{w} ({R P}^{9}) = (1 + a)^{- 10} = \sum_{i \in S_{9}} a^{i} = 1 + a^{2} + a^{4} + a^{6},$ because $S_{9} = {0, 2, 4, 6}$ : these are exactly the integers $0 \leq i \leq 9$ whose binary digits have no overlap with $9 = 1001_{2}$ . The highest nonzero term is ${\overset{ˉ}{w}}_{6} ({R P}^{9}) = a^{6} \neq 0$ , so ${R P}^{9}$ does not immerse in $R^{9 + k}$ for $k \leq 5$ , i.e. not in $R^{14}$ , in agreement with the classical computation. For the small case ${R P}^{4}$ one gets $\overset{ˉ}{w} ({R P}^{4}) = (1 + a)^{- 5} = 1 + a + a^{2} + a^{3}$ , so ${R P}^{4}$ does not immerse in $R^{5}$ or $R^{6}$ .

Facts & Assumptions

Given: The rings $H^{*} ({R P}^{m}; F_{2}) = F_{2} [a] / (a^{m + 1})$ for $m = 9$ and $m = 4$ , and AC.

[F1]

For every $m \geq 1$ the normal total class is $\overset{ˉ}{w} ({R P}^{m}) = (1 + a)^{- (m + 1)} = \sum_{i \in S_{m}} a^{i}$ with $S_{m} = {0 \leq i \leq m : i \land m = 0}$ and digitwise AND, and the highest nonzero term is ${\overset{ˉ}{w}}_{d (m)} = a^{d (m)}$ with $d (m) = \max S_{m}$ ; ${R P}^{m}$ then does not immerse in $R^{m + k}$ for any $k < d (m)$ (Real projective space Stiefel-Whitney non-immersion obstruction, The inverse of one plus the generator in the truncated mod-two polynomial ring, Normal Stiefel-Whitney and Pontryagin classes of a closed manifold).

[F2]

For the trivial rank- $(m + 1)$ bundle over the one-point base the projective-bundle theorem gives $P (ε^{m + 1}) = {R P}^{m}$ and the ring $H^{*} ({R P}^{m}; F_{2}) = F_{2} [a] / (a^{m + 1})$ free on $1, a, \dots, a^{m}$ , the relation classes $c_{i} \in H^{i} (p t; F_{2})$ vanishing for $i \geq 1$ by the dimension axiom for singular cohomology; hence the powers $1, a, \dots, a^{m}$ are linearly independent, so a coefficient displayed as $1$ gives a nonzero class (Mod-two real projective bundle theorem, Singular cohomology satisfies the Eilenberg Steenrod cohomology axioms, Real projective bundle and tautological line). AC is the hypothesis of the suppliers (The Axiom of Choice).

Verification

technique · direct

1.1F1

For $m = 9 = 1001_{2}$ the condition $i \land 9 = 0$ for $0 \leq i \leq 9$ excludes exactly the binary digit positions $0$ and $3$ , so $i$ ranges over the numbers with digits only among positions $1$ and $2$ , that is $i \in {0, 2, 4, 6}$ ; hence $S_{9} = {0, 2, 4, 6}$ and, by [F1], $\overset{ˉ}{w} ({R P}^{9}) = (1 + a)^{- 10} = 1 + a^{2} + a^{4} + a^{6} .$

2.1F1F2step 1.1

The top nonzero term is ${\overset{ˉ}{w}}_{6} ({R P}^{9}) = a^{6}$ , nonzero by [F2]; hence $d (9) = 6$ and [F1] forbids an immersion of ${R P}^{9}$ into $R^{9 + k}$ for every $k < 6$ , in particular $k = 5$ : there is no immersion into $R^{14}$ .

3.1F1F2step 1.1∎

For $m = 4 = 100_{2}$ the condition $i \land 4 = 0$ for $0 \leq i \leq 4$ allows exactly $i \in {0, 1, 2, 3}$ , so $S_{4} = {0, 1, 2, 3}$ , $d (4) = 3$ , and $\overset{ˉ}{w} ({R P}^{4}) = (1 + a)^{- 5} = 1 + a + a^{2} + a^{3} .$ The top nonzero term is ${\overset{ˉ}{w}}_{3} = a^{3} \neq 0$ , so ${R P}^{4}$ does not immerse in $R^{4 + k}$ for $k < 3$ , in particular not in $R^{5}$ or $R^{6}$ . The computations concern the class calculations only; no assertion is made about higher-codimension immersions or about embeddings.

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Sources

John W. Milnor and James D. Stasheff, Characteristic Classes (Annals of Mathematics Studies 74, Princeton University Press; complete text) (standard reference, not scraped)
Ralph L. Cohen, Immersions of Manifolds and Homotopy Theory (lecture notes, 30 June 2022; complete 46-page text) (standard reference, not scraped)
Arkadiy Skopenkov, Embedding and Knotting of Manifolds in Euclidean Spaces (arXiv:math/0604045) (standard reference, not scraped)