Talk:Exact sequence

Link

Someone reverted the addition of:

--External links-- Short Exact Sequences, explanation by Matthew Salomone

I think that's a shame because it gives a much better explanation than anything contained in the article, which is not very well written. Perhaps it should be restored?

Stikko (talk) 21:44, 26 September 2021 (UTC)

Please, read WP:EL, and specifically the first item of WP:ELNO (One should generally avoid providing external links to [...] any site that does not provide a unique resource beyond what the article would contain if it became a featured article. In other words, the site should not merely repeat information that is already or should be in the article). This applies to this external link. Also WP:NOR applies to this video, which, in any case is not a reliable source for Wikipedia. More specifically, in mathematics, YouTube videos are generally not accepted, except in very exceptional cases. Instead of trying to link this YouTube video, I suggest you to use it for proposing here specific improvements to the article. D.Lazard (talk) 09:20, 27 September 2021 (UTC)

Properties

At the beginning of the section, there is the claim that "for non-commutative groups, this is the semidirect product". It seems straight up incorrect. I do not think semidirect products are any kind of products for starters: One does not have uniqueness without a pre-specified homomorphism $\phi :H\to Aut(K)$. I could still be missing something, but the amount of clarification is rather inadequate. Yeetcode (talk) 05:16, 24 November 2023 (UTC)

 Fixed D.Lazard (talk) 09:02, 24 November 2023 (UTC)

Article desperately needs to say when it assumes groups are abelian.

In numerous places — in particular, where the article discusses a split exact sequence — the article assumes groups are abelian without any mention of this assumption.

I hope someone knowledgeable about this subject can fix this.

Misleading illustration

No doubt the "Illustration of an exact sequence of groups using Euler diagrams" shows something, but I see it as more misleaing than helpful.

Because, it appears to show the image of each map as being exactly zero, rather than equal to the kernel of the next map.

Any illustration that needs an explanation of why it appears to show the wrong thing is not a helppful one.

Restore the Div/grad/curl Helmholtz decomposition example

This section is correct mathematics, it's just the helmholtz decomposition written as an exact sequence. It is not at all too advanced for this article. Vector calculus is typically less advanced than the category theory used here. It only needs minor stylistic changes to be fine.

Why was it removed? Darcourse do you have a better example of a non-trivial sequence?

Copied here:

Discussed section

Grad, curl and div in differential geometry

Another example can be derived from differential geometry, especially relevant for work on the Maxwell equations.

Consider the Hilbert space $L^{2}$ of scalar-valued square-integrable functions on three dimensions $\left\lbrace f:\mathbb {R} ^{3}\to \mathbb {R} \right\rbrace$. Taking the gradient of a function $f\in \mathbb {H} _{1}$ moves us to a subset of $\mathbb {H} _{3}$, the space of vector valued, still square-integrable functions on the same domain $\left\lbrace f:\mathbb {R} ^{3}\to \mathbb {R} ^{3}\right\rbrace$ — specifically, the set of such functions that represent conservative vector fields. (The generalized Stokes' theorem has preserved integrability.)

First, note the curl of all such fields is zero — since

$\operatorname {curl} (\operatorname {grad} f)\equiv \nabla \times (\nabla f)=0$

for all such f. However, this only proves that the image of the gradient is a subset of the kernel of the curl. To prove that they are in fact the same set, prove the converse: that if the curl of a vector field ${\vec {F}}$ is 0, then ${\vec {F}}$ is the gradient of some scalar function. This follows almost immediately from Stokes' theorem (see the proof at conservative force.) The image of the gradient is then precisely the kernel of the curl, and so we can then take the curl to be our next morphism, taking us again to a (different) subset of $\mathbb {H} _{3}$.

Similarly, we note that

$\operatorname {div} \left(\operatorname {curl} {\vec {v}}\right)\equiv \nabla \cdot \nabla \times {\vec {v}}=0,$

so the image of the curl is a subset of the kernel of the divergence. The converse is somewhat involved (for the general case see Poincaré lemma):

Proof that $\operatorname {div} {\vec {F}}$ = 0 implies ${\vec {F}}=\operatorname {curl} {\vec {A}}$ for some ${\vec {A}}$

We shall proceed by construction: given a vector field ${\vec {F}}=F_{x}{\hat {i}}+F_{y}{\hat {j}}+F_{z}{\hat {k}}$ such that $\nabla \cdot {\vec {F}}=0$, we produce a field ${\vec {A}}$ such that ${\begin{aligned}\nabla \times {\vec {A}}&=\left(\partial _{y}A_{z}-\partial _{z}A_{y}\right){\hat {i}}+\left(\partial _{z}A_{x}-\partial _{x}A_{z}\right){\hat {j}}+\left(\partial _{x}A_{y}-\partial _{y}A_{x}\right){\hat {k}}\\&=F_{x}{\hat {i}}+F_{y}{\hat {j}}+F_{z}{\hat {k}}={\vec {F}}\end{aligned}}$ First, note that since as proved above $\operatorname {curl} (\operatorname {grad} f)=0$, we can add the gradient of any scalar function to ${\vec {A}}$ without changing the curl. We can use this gauge freedom to set any one component of ${\vec {A}}$ to zero without changing its curl; choosing arbitrarily the z-component, we thus require simply that $-\partial _{z}A_{y}{\hat {i}}+\partial _{z}A_{x}{\hat {j}}+\left(\partial _{x}A_{y}-\partial _{y}A_{x}\right){\hat {k}}=F_{x}{\hat {i}}+F_{y}{\hat {j}}+F_{z}{\hat {k}}$ Then by simply integrating the first two components, and noting that the 'constant' of integration may still depend on any variable not integrated over, we find that ${\begin{aligned}A_{x}&=\int F_{y}dz+{\vec {C_{1}}}(x,y)\\A_{y}&=-\int F_{x}dz+{\vec {C_{2}}}(x,y)\end{aligned}}$ Note that since the two integration terms ${\vec {C_{1}}},{\vec {C_{2}}}$ both depend only on x and y and not on z, then we can add another gradient of some function $f(x,y)$ that also does not depend on z. This permits us to eliminate either of the terms in favor of the other, without spoiling our earlier work that set $A_{z}$ to zero. Choosing to eliminate ${\vec {C_{1}}}$ and applying the last component as a constraint, we have ${\begin{aligned}-\partial _{x}\int F_{x}dz+\partial _{x}{\vec {C_{2}}}(x,y)-\partial _{y}\int F_{y}dz&=F_{z}\\-\int \left(\partial _{x}F_{x}+\partial _{y}F_{y}\right)dz+\partial _{x}{\vec {C_{2}}}(x,y)&=F_{z}\end{aligned}}$ By assumption, $\operatorname {div} {\vec {F}}=\partial _{x}F_{x}+\partial _{y}F_{y}+\partial _{z}F_{z}=0$, and so $\int \partial _{z}F_{z}dz+\partial _{x}{\vec {C_{2}}}(x,y)=F_{z}$ Since the fundamental theorem of calculus requires that the first term above be precisely $F_{z}$ plus a constant in z, a solution to the above system of equations is guaranteed to exist.

Having thus proved that the image of the curl is precisely the kernel of the divergence, this morphism in turn takes us back to the space we started from $L^{2}$. Since definitionally we have landed on a space of integrable functions, any such function can (at least formally) be integrated in order to produce a vector field which divergence is that function — so the image of the divergence is the entirety of $L^{2}$, and we can complete our sequence:

$0\to L^{2}\mathrel {\xrightarrow {\operatorname {grad} } } \mathbb {H} _{3}\mathrel {\xrightarrow {\operatorname {curl} } } \mathbb {H} _{3}\mathrel {\xrightarrow {\operatorname {div} } } L^{2}\to 0$

Equivalently, we could have reasoned in reverse: in a simply connected space, a curl-free vector field (a field in the kernel of the curl) can always be written as a gradient of a scalar function (and thus is in the image of the gradient). Similarly, a solenoidal vector field can be written as a curl of another field. (Reasoning in this direction thus makes use of the fact that 3-dimensional space is topologically trivial.)

This short exact sequence also permits a much shorter proof of the validity of the Helmholtz decomposition that does not rely on brute-force vector calculus. Consider the subsequence

$0\to L^{2}\mathrel {\xrightarrow {\operatorname {grad} } } \mathbb {H} _{3}\mathrel {\xrightarrow {\operatorname {curl} } } \operatorname {im} (\operatorname {curl} )\to 0.$

Since the divergence of the gradient is the Laplacian, and since the Hilbert space of square-integrable functions can be spanned by the eigenfunctions of the Laplacian, we already see that some inverse mapping $\nabla ^{-1}:\mathbb {H} _{3}\to L^{2}$ must exist. To explicitly construct such an inverse, we can start from the definition of the vector Laplacian

$\nabla ^{2}{\vec {A}}=\nabla \left(\nabla \cdot {\vec {A}}\right)+\nabla \times \left(\nabla \times {\vec {A}}\right)$

Since we are trying to construct an identity mapping by composing some function with the gradient, we know that in our case $\nabla \times {\vec {A}}=\operatorname {curl} \left({\vec {A}}\right)=0$. Then if we take the divergence of both sides

${\begin{aligned}\nabla \cdot \nabla ^{2}{\vec {A}}&=\nabla \cdot \nabla \left(\nabla \cdot {\vec {A}}\right)\\&=\nabla ^{2}\left(\nabla \cdot {\vec {A}}\right)\\\end{aligned}}$

we see that if a function is an eigenfunction of the vector Laplacian, its divergence must be an eigenfunction of the scalar Laplacian with the same eigenvalue. Then we can build our inverse function $\nabla ^{-1}$ simply by breaking any function in $\mathbb {H} _{3}$ into the vector-Laplacian eigenbasis, scaling each by the inverse of their eigenvalue, and taking the divergence; the action of $\nabla ^{-1}\circ \nabla$ is thus clearly the identity. Thus by the splitting lemma,

$\mathbb {H} _{3}\cong L^{2}\oplus \operatorname {im} (\operatorname {curl} )$,

or equivalently, any square-integrable vector field on $\mathbb {R} ^{3}$ can be broken into the sum of a gradient and a curl — which is what we set out to prove.

References

ComeAlongBort (talk) 00:09, 20 March 2025 (UTC)

I agree with the revert. This section does not belong to this article. Indeed, this section is about the interpretation of Helmholtz decomposition in tems of an exact sequence. As it is, it is not understandable for most readers of this article, since it involve advanced knowledge on differential geometry, while most readers are presumably interested in algebra, and more specifically, homological algebra. It is also not really useful for people interested in Helmholtz decomposition, since it is not here that they are supposed to find this information.

However, one sentence on this subject could be acceptable in section § Applications of exact sequences if this interpretation would be already given elsewhere in Wikipedia (this seems not to be the case).

You ask for better non-trivial examples. Here are a few: free resolutions, homological dimension, Hilbert's syzygy theorem, and more generally the whole homological algebra. D.Lazard (talk) 21:06, 20 March 2025 (UTC)