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For example a calculation of $S^5\overset{\eta}{\to} S^4\overset{2}{\to} S^4\overset{\eta}{\to} S^3$.

I know this exists in Toda's book. However, I'm looking for a fairly elementary proof that is easy to insert as an example into a paper (although I'll be happy for any reference).

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    $\begingroup$ Did you see Neil Strickland's comment here: mathoverflow.net/a/173063/16785? I think it's also in Mosher and Tangora 'Cohomology Operations...'. Also Kochman: 'Bordism, Stable Homotopy and Adams Spectral Sequences' $\endgroup$ – Drew Heard Nov 20 '19 at 15:44
  • $\begingroup$ @DrewHeard I saw his comment, but it lacks proof... I think Tyler's answer is similar to the method in Mosher and Tangora, just hides somewhat the spectral sequence involved (which is good for my purposes). $\endgroup$ – JK34 Nov 21 '19 at 14:27
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There is an argument that this Toda bracket does not contain zero. However, it uses the relationship between vanishing Toda brackets and the ability to iteratively factor through mapping cones. Let's sketch that first.


Suppose we have a triple composite $X \xrightarrow{f} Y \xrightarrow{g} Z \xrightarrow{h} W$. If $gf$ is nullhomotopic and $hg$ is nullhomotopic, then there exists nullhomotopies: paths $K: [0,1] \to Map_*(X,Z)$ from $gf$ to the trivial map, and $L:[0,1] \to Map_*(Y,W)$ from $hg$ to the trivial map. Gluing together $h \circ K$ and $L \circ f$, we get a map $S^1 \to Map_*(X,W)$, or $\Sigma X \to W$: a representative for the Toda bracket. Choices of $K$ and $L$ change this element. If the Toda bracket contained zero, then we could make choices of $K$ and $L$ so that this map is trivial: it extends to a map $D^2 \to Map_*(X,W)$.

If this is true, then we can use this data to build a space. The nullhomotopy $K$ of $gf$ allows us to factor $g$ through a map $k: Cf \to Z$ from the mapping cone. The nullhomotopy $L$ allows us to factor $h$ through a map $\ell: Cg \to W$. The map from $D^2$ above allows more: it allows us to factor the map $\ell$ through a map $p: Ck \to W$ from the mapping cone. And then we can take the mapping cone on $p$: a space $A$. This space $A$ would live in a diagram of homotopy pushouts:

iterated mapping cones


Let's we apply this to the composite $S^5 \to S^4 \to S^4 \to S^3$ you described. First, we get a map $k: C\eta \to S^4$ (the source has a 4-dimensional cell and a 6-dimensional cell). Then, extending over the mapping cone, we get a map $p: Ck \to S^3$ (the source has a 4-dimensional cell, a 5-dimensional cell, and a 7-dimensional cell). Finally, we take the cone on $p$, getting a CW-complex $A$ with cells in dimensions 3, 5, 6, and 8. The attaching maps for the cells are related as well: $\eta$ is the attaching map for the 5-cell to the 3-cell; $2$ is the attaching map for the 6-cell to the 5-cell; $\eta$ is the attaching map for the $8$-cell to the $6$-cell.

Moreover, on cohomology we'd find that $H^*(A)$ is $\Bbb Z/2$ in dimensions 3, 5, 6, and 8 (one for each cell), and $0$ elsewhere. The attaching maps are also detected by Steenrod operations: $Sq^2$ takes the 3-dimensional generator to the 5-dimensional generator; $Sq^1$ takes the 5-dimensional generator to the $6$-dimensional generator; and $Sq^2$ takes the $6$-dimensional generator to the $8$-dimensional one.

However, $$ Sq^2 Sq^1 Sq^2 = Sq^2 Sq^3 = Sq^5 + Sq^4 Sq^1 = Sq^1 Sq^4 + Sq^4 Sq^1 $$ and the latter operation, applied to the generator in degree 3, must be zero because all the terms factor through the zero groups $H^4(A)$ and $H^7(A)$.

Let's say we know $\pi_6(S^3)$, the home for this Toda bracket. It is $\Bbb Z/12$: it is a subgroup of $\Bbb Z/24 \cdot \{\nu\}$, the third stable homotopy group of spheres. The bracket involves 2-torsion homotopy groups and so it has to live in the 2-torsion part $\Bbb Z/4$. The indeterminacy in the bracket is the image of the Hopf map $\eta$: this is the subgroup $\Bbb Z/2$. So this leaves us little choice: the bracket $\langle \eta, 2, \eta\rangle$ is $2\nu$, up to indeterminacy.


We note that $Sq^1$ detects $2$, $Sq^2$ detects $\eta$, and $Sq^4$ detects $\nu$. It is no accident, given this, that $\langle \eta, 2, \eta\rangle = 2 \nu$ at the same time that we have the Adem relation $Sq^2 Sq^1 Sq^2 = Sq^4 Sq^1 + Sq^1 Sq^4$. Relations between Steenrod operations detect brackets in the homotopy groups of spheres (and, conversely, relations between elements in the homotopy groups of spheres detect brackets between Steenrod operations).

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  • $\begingroup$ This is exactly what I was looking for, thank you! I would just like a source for your statement "The attaching maps are also detected by Steenrod operations". I'm only aware of this phenomenon for the hopf map in 4-polyhedra ... $\endgroup$ – JK34 Nov 21 '19 at 14:36
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    $\begingroup$ Sure. So the mapping cone of $2: S^1 \to S^1$ is $\Bbb{RP}^2$, and $Sq^1$ takes the 1-dimensional cohomology generator of $\Bbb{RP}^2$ to the 2-dimensional one. Suspension preserves all this, so the mapping cone of $2: S^n \to S^n$ is $\Sigma^{n-1} \Bbb{RP}^2$: Steenrod squares are preserved by suspension, and so we still have $Sq^1$. Similarly, the mapping cone of $\eta: S^3 \to S^2$ is $\Bbb{CP}^2$, and the operation $Sq^2$ in $H^*(\Bbb{CP}^2)$ takes the 2-dimensional generator to the 4-dimensional generator, and similarly for the suspensions of $\eta$. Does that answer your question? $\endgroup$ – Tyler Lawson Nov 21 '19 at 19:08
  • $\begingroup$ @Tylor No, I meant something like a book where the precise meaning of that sentence is explained (I understand you are implying in your comment that it's because they are EM spaces, but I need a source to cite). $\endgroup$ – JK34 Nov 21 '19 at 20:08
  • $\begingroup$ @JK34 I guess one reference might be section 3 of J.F. Adams' paper "On the groups J(X) IV": he uses this to show that a map f:X -> Y that induces 0 on homology determines an invariant, e(f), in Ext over the Steenrod algebra; he attributes the approach to Steenrod, and discusses how it is related to the Sq^n. (J.P.C. Greenlees also has an expository paper called "How blind is your favourite cohomology theory?" which I enjoyed back in graduate school: it discusses this at a milder pace, but it is a little harder to get.) $\endgroup$ – Tyler Lawson Nov 21 '19 at 21:06
  • $\begingroup$ Is this your original proof? I suppose I should credit you/cite the place where you took it from... $\endgroup$ – JK34 Nov 22 '19 at 20:35

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