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• CommentRowNumber1.
• CommentAuthorUrs
• CommentTimeMar 23rd 2019

am starting an entry here in order to record some facts. Not done yet

• CommentRowNumber2.
• CommentAuthorUrs
• CommentTimeMar 23rd 2019
• (edited Mar 23rd 2019)

added statement of this fact (here)

Let $X$ be a closed connected 8-manifold. Then $X$ has G-structure for $G =$ Spin(5) if and only if the following conditions are satisfied:

1. The second and sixth Stiefel-Whitney classes (of the tangent bundle) vanish

$w_2 \;=\; 0 \phantom{AAA} w_6 \;=\; 0$
2. The Euler class $\chi$ (of the tangent bundle) evaluated on $X$ (hence the Euler characteristic of $X$) is proportional to I8 evaluated on $X$:

\begin{aligned} 8 \chi[X] &= 192 \cdot I_8[X] \\ & = 4 \Big( p_2 - \tfrac{1}{2}\big(p_1\big)^2 \Big)[X] \end{aligned}
3. The Euler characteristic is divisible by 4:

$\chi[X] \;=\; 0 \;\in\; \mathbb{Z}/4$
• CommentRowNumber3.
• CommentAuthorUrs
• CommentTimeMar 23rd 2019
• (edited Mar 23rd 2019)

added statement of this fact (here):

Let $X$ be a closed connected spin 8-manifold. Then $X$ has G-structure for $G =$ Spin(4)

$\array{ && B Spin(4) \\ & {}^{\mathllap{ \widehat{T X} }} \nearrow & \big\downarrow \\ X & \underset{T X}{\longrightarrow} & B Spin(8) }$

if and only if the following conditions are satisfied:

1. the sixth Stiefel-Whitney class of the tangent bundle vanishes

$w_6(T X) \;=\; 0$
2. the Euler class of the tangent bundle vanishes

$\chi_8(T X) \;=\; 0$
3. the I8-term evaluated on $X$ is divisible as:

$\tfrac{1}{32} \Big( p_2 - \big( \tfrac{1}{2} \big( p_1 \big)^2 \big) \Big) \;\in\; \mathbb{Z}$
4. there exists an integer $k \in \mathbb{Z}$ such that

1. $p_2 = (2k - 1)^2 \left( \tfrac{1}{2} p_1 \right)^2$;

2. $\tfrac{1}{3} k (k+2) p_2[X] \;\in\; \mathbb{Z}$.

Moreover, in this case we have for $\widehat T X$ a given Spin(4)-structure as in (eq:Spin4Structure) and setting

$\widetilde G_4 \;\coloneqq\; \tfrac{1}{2} \chi_4(\widehat{T X}) + \tfrac{1}{4}p_1(T X)$

for $\chi_4$ the Euler class on $B Spin(4)$ (which is an integral class, by this Prop.)

the following relations:

1. $\tilde G_4$ (eq:TildeG4) is an integer multiple of the first fractional Pontryagin class by the factor $k$ from above:

$\widetilde G_4 \;=\; k \cdot \tfrac{1}{2}p_1$
2. The (mod-2 reduction followed by) the Steenrod operation $Sq^2$ on $\widetilde G_4$ (eq:TildeG4) vanishes:

$Sq^2 \left( \widetilde G_4 \right) \;=\; 0$
3. the shifted square of $\tilde G_4$ (eq:TildeG4) evaluated on $X$ is a multiple of 8:

$\tfrac{1}{8} \left( \left( \widetilde G_4 \right)^2 - \widetilde G_4 \big( \tfrac{1}{2} p_1\big)[X] \right) \;\in\; \mathbb{Z}$
4. The I8-term is related to the shifted square of $\widetilde G_4$ by

$$4 \Big( \left( \widetilde G_4 \right)^2 - \widetilde G_4 \left( \tfrac{1}{2}p_1 \right) \Big) \;=\; \Big( p_2 • \big( \tfrac{1}{2}p_1 \big)^2 \Big)$$
• CommentRowNumber4.
• CommentAuthorUrs
• CommentTimeMar 23rd 2019
• (edited Mar 23rd 2019)

added a brief mentioning of 8-manifolds with exotic boundary 7-spheres (here) – so far just a glorified pointer to

1. Note that the signature of an 8-manifold need not necessarily be 1 or -1. Take for example the spheres S^{4k}, k a positive integer, which has signature 0.

Cole Durham

• CommentRowNumber6.
• CommentAuthorUrs
• CommentTimeSep 9th 2020
• (edited Sep 9th 2020)

Thanks for catching.

(This statement was copy-and-pasted from discussion of Milnor’s construction of exotic 7-spheres, where the signature is $\pm 1$.)

So I have fixed the actual formula now.