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I'm starting to understand what commutative diagrams are, but I'm not sure about their purpose, what is their intended use and what kind of problems are solvable with them. By "solvable with a commutative diagram" I mean some fancy graphical reasoning, redrawing etc.

For example given only the commutative diagram for the exterior derivative

$$ \newcommand{\ra}[1]{\kern-1.5ex\xrightarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex} \newcommand{\ras}[1]{\kern-1.5ex\xrightarrow{\ \ \smash{#1}\ \ }\phantom{}\kern-1.5ex} \newcommand{\da}[1]{\bigg\downarrow\raise.5ex\rlap{\scriptstyle#1}} \begin{array}{c} \Omega^k(N) & \ra{f^*} & \Omega^k(M) \\ \da{d} & & \da{d} \\ \Omega^{k+1}(N) & \ras{f^*} & \Omega^{k+1}(M) \\ \end{array} $$

is it even possible to tell that $d$ is a derivative, that is it is linear and the appropriate Leibniz rule holds?

Another example is my own, I may have done it totally wrong.

Given two vector spaces $V$ and $W$ (possibly of the same dimension) with scalar products, $f$ being a morphism, is it possible to prove that the following diagram commutes:

$$ \newcommand{\ra}[1]{\kern-1.5ex\xrightarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex} \newcommand{\ras}[1]{\kern-1.5ex\xrightarrow{\ \ \smash{#1}\ \ }\phantom{}\kern-1.5ex} \newcommand{\da}[1]{\bigg\downarrow\raise.5ex\rlap{\scriptstyle#1}} \begin{array}{c} V & \ra{f} & W \\ \da{h} & & \da{h} \\ V & \ras{f} & W \\ \end{array} $$

only for $h$ of a certain form, which I guess is

$$h(v) = \lambda(v^2) v$$

where $\lambda$ is some arbitrary function? Is it a valid commutative diagram?

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7  
How can $h$ be a map $V\to V$ and at the same time $W\to W$? –  wildildildlife Oct 14 '12 at 19:27
    
Thanks for showing how to make commutative diagrams here :) –  Yuri Vyatkin Oct 14 '12 at 23:37
    
@wildildildlife I indeed had an uneasy feeling about that while writing the diagram, but then I thought it should bare the same meaning as $d$ being of types $\Omega^k(N) \to \Omega^{k+1}(N)$ and $\Omega^k(M) \to \Omega^{k+1}(M)$ simultaneously. Isn't it valid given $V$ and $W$ are both real vector spaces? –  Yrogirg Oct 15 '12 at 6:55
    
Operator $d$ is a natural mapping (its definition is valid for all manifolds and does not require any additional choice), but your $h$ depends on a choice of the inner product in vector spaces, so it cannot be natural. –  Yuri Vyatkin Oct 15 '12 at 9:46
    
@YuriVyatkin but I meant not just vector spaces, but vector spaces with scalar products, I don't know the name for such structures, possibly Euclidean space? So $f$ is not just a linear map, but an orthogonal map. –  Yrogirg Oct 15 '12 at 9:56

2 Answers 2

up vote 17 down vote accepted

Commutative diagrams aren't a problem solving technique in the usual sense of the term. Rather, they're a language for efficiently describing certain kinds of relationships. You could remove the commutative diagrams from any proof that used them, but it would make the proof longer and harder to understand. (You could remove the commutative diagrams from the statements of any theorems that used them, too, but those would also become longer and harder to understand.)

If you haven't run into a problem where it seems like using a commutative diagram would help you express some idea, then I wouldn't try to force the issue. Wait until you actually feel the need.

In this particular case, the first commutative diagram you wrote down expresses the naturality of the exterior derivative. This is a useful thing to know: it guarantees that any computations you do involving exterior derivatives remain valid under pullback. But if you haven't run across a situation where it would be useful to know this, then wait.

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I agree on both counts. I might liken them to maps, tables, and graphs: they’re all ways of presenting particular kinds of information effectively. –  Brian M. Scott Oct 14 '12 at 23:17
    
You meant naturality, right? –  Alexei Averchenko Oct 16 '12 at 9:25
    
@Alexei: yes, thank you. –  Qiaochu Yuan Oct 16 '12 at 18:12

A commutative diagram is a statement. In the case of your last diagram its claims that for all $v \in V$ $$ f(h_V (v)) = h_W (f(v)) \tag{1} $$ where $h_{V} \colon V \to V$ is defined as $$h_V (v) = \lambda ( \langle v,v \rangle_V ) v $$ where $\lambda$ is an arbitrary function $\lambda \colon \Bbb R \to \Bbb R$.

Substituting the definition of $h$ into (1) we have $$ f\Big(\lambda( \langle v,v \rangle_V) v\Big) = \lambda (\langle f(v),f(v) \rangle_W) f(v) \tag{2} $$ Assuming that f is an isometry, that is f is linear and satisfies $$\langle f(v),f(v) \rangle_W = \langle v,v \rangle_V $$ so (2) becomes $$ \lambda \langle v,v \rangle_V f(v) = \lambda \langle v,v \rangle_V f(v) $$

Since we require $v$ be an arbitrary element of $V$ we may conclude that $$ f(v) = f(v) $$ for all $v \in V$.

Thus, the statement of your second diagram is true.

Edit. I corrected the above calculation to show that $\lambda$ can be an arbitrary real function, not just a multiplication to a scalar, as I erroneously assumed initially.

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Thank you for your replay. But do you mean the commutative diagram does not hold for $h(v) = \langle v, v \rangle^2 v$ or $h(v) = \sin(\langle v, v \rangle) v$? By lambda I meant $\mathbb R \to \mathbb R$, that's why I wrote $\lambda(v^2)$ "some arbitrary function". Anyway, the whole point of the question was whether I can deduce the form of $h$ with some graphic technique involving commutative diagrams. If no, then no. For the usual analytic approach there are two articles, first one is springerlink.com/index/X88326542Q707535.pdf The results are quite of use in fluid dynamics. –  Yrogirg Oct 16 '12 at 7:29
    
the link above is a bit slow, it needs some time to show the preview. –  Yrogirg Oct 16 '12 at 7:30
1  
@Yrogirg It looks to me that I was in rush this morning and in fact my simple calculation worked for any real function $\lambda$. And of course, there is nothing to do with commutative diagrams: they are just a notation! –  Yuri Vyatkin Oct 16 '12 at 9:13

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