The attaching map for the top cell of the torus $S^n \times S^n$ is a map $$ [x,y]: S^{2n-1} \to S^n \vee S^n $$ where the notation is such that $x,y : S^n \to S^n \vee S^n$ are the two inclusions–––the map $[x,y]$ is the generalized Whitehead product of $x$ and $y$. (More generally, if $f: \Sigma X \to \Sigma Z$ and $g: \Sigma Y \to \Sigma Z$ are maps, we have a map $[f,g]: \Sigma (X\wedge Y) \to \Sigma Z$.)

It is not difficult to show that this map has a framed manifold description via the Pontryagin construction: Consider the standard inclusion $$ S^{n-1} \times D^n \subset \partial (D^n \times D^n) = S^{2n-1} $$ Then $P = S^{n-1}\times 0$ and $Q:= \ast \times S^{n-1}$ (where $\ast$ is the basepoint of $S^{n-1}$) are a pair of disjoint framed manifolds in $S^{2n-1}$ (with trivial framings in each case) having linking number one. Then a version of Pontryagin construction applied to $P \amalg Q \subset S^{2n-1}$ defines the map $[x,y]$. (The map is given by sending a point in a tubular neighborhood of $P$ to the first sphere and a point in a tubular neighborhood of $Q$ to the second sphere in each case using the Pontryagin construction.)

I am really interested in finding an analogous description for iterated Whitehead products: for example there is a map $$ [[x,y],z]: S^{3n-2} \to S^n \vee S^n \vee S^n $$ where $x,y,z$ are the three inclusions of $S^n$ in the three fold wedge. I'd like to have a framed manifold description of this map. Presumably, there should be a "link" $$ S^{n-1} \amalg S^{n-1} \amalg S^{n-1} \subset S^{3n-2} $$ where each component is representing the trivial framed bordism class. This link should represent the map $[[x,y],z]$ via the Pontryagin construction.

How does this work?

The attaching map for the top cell of the torus $S^n \times S^n$ is a map $$ [x,y]: S^{2n-1} \to S^n \vee S^n $$ where the notation is such that $x,y : S^n \to S^n \vee S^n$ are the two inclusions–––the map $[x,y]$ is the generalized Whitehead product of $x$ and $y$. (More generally, if $f: \Sigma X \to \Sigma Z$ and $g: \Sigma Y \to \Sigma Z$ are maps, we have a map $[f,g]: \Sigma (X\wedge Y) \to \Sigma Z$.)

It is not difficult to show that this map has a framed manifold description via the Pontryagin construction: Consider the standard inclusion $$ S^{n-1} \times D^n \subset \partial (D^n \times D^n) = S^{2n-1} $$ Then $P = S^{n-1}\times 0$ and $Q:= \ast \times S^{n-1}$ (where $\ast$ is the basepoint of $S^{n-1}$) are a pair of disjoint framed manifolds in $S^{2n-1}$ (with trivial framings in each case) having linking number one. Then a version of Pontryagin construction applied to $P \amalg Q \subset S^{2n-1}$ defines the map $[x,y]$. (The map is given by sending a point in a tubular neighborhood of $P$ to the first sphere and a point in a tubular neighborhood of $Q$ to the second sphere in each case using the Pontryagin construction.)

I am really interested in finding an analogous description for iterated Whitehead products: for example there is a map $$ [[x,y],z]: S^{3n-2} \to S^n \vee S^n \vee S^n $$ where $x,y,z$ are the three inclusions of $S^n$ in the three fold wedge. I'd like to have a framed manifold description of this map. Presumably, there should be a "link" $$ S^{2n-2} \amalg S^{2n-2} \amalg S^{2n-2} \subset S^{3n-2} $$ where each component is representing the trivial framed bordism class. This link should represent the map $[[x,y],z]$ via the Pontryagin construction.

How does this work?

The attaching map for the top cell of the torus $S^n \times S^n$ is a map $$ [x,y]: S^{2n-1} \to S^n \vee S^n $$ where the notation is such that $x,y : S^n \to S^n \vee S^n$ are the two inclusions–––the map $[x,y]$ is the generalized Whitehead product of $x$ and $y$. (More generally, if $f: \Sigma X \to \Sigma Z$ and $g: \Sigma Y \to \Sigma Z$ are maps, we have a map $[f,g]: \Sigma (X\wedge Y) \to \Sigma Z$.)

It is not difficult to show that this map has a framed manifold description via the Pontryagin-Thom construction: Consider the standard inclusion $$ S^{n-1} \times D^n \subset \partial (D^n \times D^n) = S^{2n-1} $$ Then $P = S^{n-1}\times 0$ and $Q:= \ast \times S^{n-1}$ (where $\ast$ is the basepoint of $S^{n-1}$) are a pair of disjoint framed manifolds (trivial framings in each case) having linking number one. Then a version of Pontryagin construction applied to $P \amalg Q \subset S^{2n-1}$ defines the map $[x,y]$. (The map is given by sending a point in a tubular neighborhood of $P$ to the first sphere and a point in a tubular neighborhood of $Q$ to the second sphere in each case using the Pontryagin construction).

I am really interested in finding an analogous description for iterated Whitehead products: for example there is a map $$ [[x,y],z]: S^{3n-2} \to S^n \vee S^n \vee S^n $$ where $x,y,z$ are the three inclusions of $S^n$ in the three fold wedge. I'd like to have a framed manifold description of this map. Presumably, there should be a "link" $$ S^{n-1} \amalg S^{n-1} \amalg S^{n-1} \subset S^{3n-2} $$ where each component is representing the trivial framed bordism class. This link should represent the map $[[x,y],z]$ via the Pontryagin construction.

How does this work?

The attaching map for the top cell of the torus $S^n \times S^n$ is a map $$ [x,y]: S^{2n-1} \to S^n \vee S^n $$ where the notation is such that $x,y : S^n \to S^n \vee S^n$ are the two inclusions–––the map $[x,y]$ is the generalized Whitehead product of $x$ and $y$. (More generally, if $f: \Sigma X \to \Sigma Z$ and $g: \Sigma Y \to \Sigma Z$ are maps, we have a map $[f,g]: \Sigma (X\wedge Y) \to \Sigma Z$.)

It is not difficult to show that this map has a framed manifold description via the Pontryagin construction: Consider the standard inclusion $$ S^{n-1} \times D^n \subset \partial (D^n \times D^n) = S^{2n-1} $$ Then $P = S^{n-1}\times 0$ and $Q:= \ast \times S^{n-1}$ (where $\ast$ is the basepoint of $S^{n-1}$) are a pair of disjoint framed manifolds in $S^{2n-1}$ (with trivial framings in each case) having linking number one. Then a version of Pontryagin construction applied to $P \amalg Q \subset S^{2n-1}$ defines the map $[x,y]$. (The map is given by sending a point in a tubular neighborhood of $P$ to the first sphere and a point in a tubular neighborhood of $Q$ to the second sphere in each case using the Pontryagin construction.)

I am really interested in finding an analogous description for iterated Whitehead products: for example there is a map $$ [[x,y],z]: S^{3n-2} \to S^n \vee S^n \vee S^n $$ where $x,y,z$ are the three inclusions of $S^n$ in the three fold wedge. I'd like to have a framed manifold description of this map. Presumably, there should be a "link" $$ S^{n-1} \amalg S^{n-1} \amalg S^{n-1} \subset S^{3n-2} $$ where each component is representing the trivial framed bordism class. This link should represent the map $[[x,y],z]$ via the Pontryagin construction.

How does this work?

The attaching map for the top cell of the torus $S^n \times S^n$ is a map $$ [x,y]: S^{2n-1} \to S^n \vee S^n $$ where the notation is such that $x,y : S^n \to S^n \vee S^n$ are the two inclusions (the map $[x,y]$ is the generalized Whitehead product of $x$ and $y$).

It is not difficult to show that this map has a framed manifold description via the Pontryagin-Thom construction: Consider the standard inclusion $$ S^{n-1} \times D^n \subset \partial (D^n \times D^n) = S^{2n-1} $$ Then $P = S^{n-1}\times 0$ and $Q:= \ast \times S^{n-1}$ (where $\ast$ is the basepoint of $S^{n-1}$) are a pair of disjoint framed manifolds (trivial framings in each case) having linking number one. Then a version of Pontryagin construction applied to $P \amalg Q \subset S^{2n-1}$ defines the map $[x,y]$. (The map is given by sending a point in a tubular neighborhood of $P$ to the first sphere and a point in a tubular neighborhood of $Q$ to the second sphere in each case using the Pontryagin construction).

I am really interested in finding an analogous description for iterated Whitehead products: for example there is a map $$ [[x,y],z]: S^{3n-2} \to S^n \vee S^n \vee S^n $$ where $x,y,z$ are the three inclusions of $S^n$ in the three fold wedge. I'd like to have a framed manifold description of this map. Presumably, there should be a "link" $$ S^{n-1} \amalg S^{n-1} \amalg S^{n-1} \subset S^{3n-2} $$ where each component is representing the trivial framed bordism class. This link should have "triple linking number one." This link should represent the map $[[x,y],z]$ via the Pontryagin construction.

How does this work?

The attaching map for the top cell of the torus $S^n \times S^n$ is a map $$ [x,y]: S^{2n-1} \to S^n \vee S^n $$ where the notation is such that $x,y : S^n \to S^n \vee S^n$ are the two inclusions–––the map $[x,y]$ is the generalized Whitehead product of $x$ and $y$. (More generally, if $f: \Sigma X \to \Sigma Z$ and $g: \Sigma Y \to \Sigma Z$ are maps, we have a map $[f,g]: \Sigma (X\wedge Y) \to \Sigma Z$.)

It is not difficult to show that this map has a framed manifold description via the Pontryagin-Thom construction: Consider the standard inclusion $$ S^{n-1} \times D^n \subset \partial (D^n \times D^n) = S^{2n-1} $$ Then $P = S^{n-1}\times 0$ and $Q:= \ast \times S^{n-1}$ (where $\ast$ is the basepoint of $S^{n-1}$) are a pair of disjoint framed manifolds (trivial framings in each case) having linking number one. Then a version of Pontryagin construction applied to $P \amalg Q \subset S^{2n-1}$ defines the map $[x,y]$. (The map is given by sending a point in a tubular neighborhood of $P$ to the first sphere and a point in a tubular neighborhood of $Q$ to the second sphere in each case using the Pontryagin construction).

I am really interested in finding an analogous description for iterated Whitehead products: for example there is a map $$ [[x,y],z]: S^{3n-2} \to S^n \vee S^n \vee S^n $$ where $x,y,z$ are the three inclusions of $S^n$ in the three fold wedge. I'd like to have a framed manifold description of this map. Presumably, there should be a "link" $$ S^{n-1} \amalg S^{n-1} \amalg S^{n-1} \subset S^{3n-2} $$ where each component is representing the trivial framed bordism class. This link should represent the map $[[x,y],z]$ via the Pontryagin construction.

How does this work?