Given a closed manifold $M^n$ and its $k$-fold product space $M^n\times\cdots\times M^n$,Can the diagonal submanifold $\Delta:=\{(m,\cdots,m)\in (M^n)^k\mid m\in M\}$ be isotopied to the submanifold
$M\times \{\ast\}\times\cdots\times \{\ast\}$?


For some particularly nice manifold,i tend to believe,the answer is positive.Say torus $T^n$.
In general,I guess,this is not true.


(Following the idea from the classification of knot theory,I think,a good understanding of the diagonal complement would help here.Say, for the $k=2$ case,the diagonal complement is a fiber bundle over $M$ with fiber $M-\{\ast\}$.But $M\times M-M\times\{\ast\}$ is always a trivial bundle.)



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> 1.Are there some more examples for $k\geq 3?$
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> 2.Have people already established some criterion for the existence of such an isotropy or any obstruction theory in terms of the algebraic topology of $M$? Are there some nice references on this topic?
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Ambient isotopy of the diagonal submanifold in product space
The answer is always no for $M$ a closed manifold. 


If $\Delta: M\to M\times M$ were isotopic into the first factor, then in particular it would be homotopic to a map $\Delta': M\to M\times M$ with image in $M\times \{ \ast\}$. Then the composition of $\Delta$ with projection onto the second factor $pr\_2\circ \Delta: M\to M$ would be homotopic to $pr\_2\circ\Delta'$, therefore null-homotopic. But this composition is the identity map, and $M$ is not contractible.


The same argument works with more factors. 


