A calculation of Pin "*" bordism groups

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PROPOSITION 10. The two tables below are obtained by smashing bo<2> with the cofibration sequence (8) and taking homotopy groups. The first table takes 2r — 1 = 4/c -h 1 and the second takes 2r — 1 = 4A: -f 3.

TCgn + i M(Z/2, 0) л bo<2> M(4Ä: 4- 1) л bo<2> I^'MiAk -f 3) л bo<2>

4

Z / 2

Z / 2

Z / 2 Ф Z/2

5

0

0

Z / 2

6

0

2 / 2 - ^ + 2

2 / 2 - ^ + 2

7

0

Z / 2

Z / 2

8

z / 2

Z / 2 Ф Z/2

Z / 2

9

Z / 2

Z / 2

0

10

Z / 4

Zj2 * " + ^

2 / 24 " + 3

11

Z / 2

Zjl

Z / 2

'tSn + i

M ( Z / 2 , 0) л bo<2>

Ы ( Ак + 3) л bo<2>

I^M { Ak + 5) л bo<2>

4

Z / 2

2Д4™ + 2

Z / 2'^ + '

5

0

Z / 2

Z / 2

6

0

z / 2

Z / 2

7

0

0

0

8

z / 2

2Д4» + 3

2 / 2 * " + 2

9

Z / 2

z / 2

Z / 2

10

Z / 4

ZI2®ZI1

Zjl® Zjl

11

Z / 2

Z / 2

Z / 2

The groups TTg^ ^, for w = 0 and / = 0 or 1 vanish for dimensional reasons. For /2=0 and i = 2 or 3, Я8„ц.,(М(4Л-h 3) л bo<2» = ^8„^,(M(Z/2,0) л bo<2» = 7r3,^,(M(4/c + l)Abo<2»=Z/2.

Proof . Tht first columns follow just as above. Indeed, ä,(M(Z/2, 0)л bo<2» = 7ü,(M(Z/2, 0) л bo<0» unless / = 0, 1 or 2, in which case 7i,(M(Z/2,0) л bo<2» = 0 if / = 0, or 1, and 7i2(M(Z/2, 0) л bo<2» = Z/2. (This is why we have started our rows with 4 and gone to 11.) From the second table bo<2> л M(3)8„ + 7 = 0 and bo<2> л M(3)8„ ^ в = ^ß- From the first table bo<2> aM(3)8„ + 4 = Z/2^" + 2 and bo<2> л M(3)8„ + 5 = ^/2. Feeding these values back into the second table, we see that bo<2> л M(3)8^4.9 has order at least 2 and at most 4. From the first table, it is a subgroup of Z/2, and so bo<2> л M(3)8„ + 9 = Z/2. From the first table, bo<2> л M(3)8„4.8 is cyclic of order at least 2*"'^^ and at most 2"*""^^. Feeding this into the second table, bo<2> л M(3)8„ + 8 = Z/2^"-^^ and bo<2> л M(3)8„+ ю has order at least 2 and at most 8. From the first table, bo<2> л M(3)8„-hii is a subgroup of Z/2, whereas