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Stable homotopy theory and generalised cohomology

Stable homotopy theory and generalised cohomology

Warm-upHuman

By the Freudenthal suspension theorem, πn+k(Sn)→πn+k+1(Sn+1)\pi_{n+k}(S^n) \to \pi_{n+k+1}(S^{n+1}) is an isomorphism when n>k+1n > k + 1. What is the smallest nn for which πn+6(Sn)\pi_{n+6}(S^n) is guaranteed to equal the stable stem π6s\pi_6^s?

Stable homotopy theory and generalised cohomology

CoreHuman

Why is complex K-theory KU∗KU^* not isomorphic to ordinary cohomology H∗(−;A)H^*(-; A) for any abelian group AA?

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Stable homotopy theory and generalised cohomology

CoreHuman

In the proof that Stiefel–Whitney numbers of a boundary M=∂WM = \partial W vanish, we wrote wI[M]=⟨i∗wI(W),∂[W,M]⟩=⟨δi∗wI(W),[W,M]⟩=0w_I[M] = \langle i^* w_I(W), \partial[W, M] \rangle = \langle \delta i^* w_I(W), [W, M] \rangle = 0. Which fact justifies the final equality?

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Stable homotopy theory and generalised cohomology

CoreHuman

Which of these closed manifolds are boundaries of compact manifolds? (Choose all that apply.)

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Stable homotopy theory and generalised cohomology

CoreHuman

The image of the J-homomorphism in π11s\pi_{11}^s is cyclic of order the denominator of B6/12B_6 / 12, where B6=142B_6 = \tfrac{1}{42}. What is this order?

Stable homotopy theory and generalised cohomology

CoreHuman

Over F3\mathbb{F}_3, what is the 33-series [3]F(x)[3]_F(x) of the multiplicative formal group law F(x,y)=x+y+xyF(x, y) = x + y + xy, and what is its height?

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Stable homotopy theory and generalised cohomology

CoreFormal

Which Lean statement formalises 'the multiplicative formal group law has height 1 at the prime pp' as it is proved in this lesson?

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Stable homotopy theory and generalised cohomology

StretchHuman

By Thom's theorem, dim⁡F2Nn\dim_{\mathbb{F}_2} \mathfrak{N}_n is the number of partitions of nn into parts not of the form 2k−12^k - 1 (that is, avoiding 1,3,7,15,…1, 3, 7, 15, \dots). What is dim⁡N6\dim \mathfrak{N}_6?

Stable homotopy theory and generalised cohomology

StretchComputational

Write image_of_j(k) returning the denominator of B2k/(4k)B_{2k}/(4k), computing the Bernoulli numbers exactly with fractions.Fraction from the recurrence ∑j=0m(m+1j)Bj=0\sum_{j=0}^{m} \binom{m+1}{j} B_j = 0 (m≥1m \ge 1, B0=1B_0 = 1). Check that it gives 24,240,504,48024, 240, 504, 480 for k=1,2,3,4k = 1, 2, 3, 4.

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