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I'm reading the Wiki article about the Grothendieck group.

What's the reason we define $[A] - [B] + [C] = 0 $ rather than $[A] + [B] - [C] = 0 $ (or something else) for every exact sequence $0 \to A \to B \to C \to 0$? What is the property we obtain if we define it this way? I suppose it has something to do with exactness at $B$ but what?

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You can also read a short exact sequence $0 \to A \to B \to C \to 0$ as "$B$ is an extension of $C$ by $A$", which can help to justify the additivity requirement explained in Pete L. Clarks answer. –  Alexander Thumm Aug 5 '12 at 19:47
@Alexander: I didn't explain anything; I only hinted. :) –  Pete L. Clark Aug 5 '12 at 19:53
@PeteL.Clark: Which makes it an even better explanation. :D –  Alexander Thumm Aug 5 '12 at 20:01
Consider the short exact sequences $0 \to A \to A \oplus C \to C \to 0$... –  Qiaochu Yuan Aug 5 '12 at 20:20

1 Answer 1

up vote 10 down vote accepted

To get a feel for this kind of relation, consider a short exact sequence

$0 \rightarrow V_1 \rightarrow V_2 \rightarrow V_3 \rightarrow 0$

of finite-dimensional vector spaces over a field. What is the relation between their dimensions?

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$\mathrm{dim} V_1 \leq \mathrm{dim} V_2$ and $\mathrm{dim} V_2 \geq \mathrm{dim} V_3$? –  Rudy the Reindeer Aug 5 '12 at 19:28
Dimension is additive on short exact sequences. –  Andrew Aug 5 '12 at 19:32
@Clark Do you remember the rank and nullity theorem from linear algebra? That's the sort of relation between the dimensions that Pete is pointing out in his answer. You can take a look here at the reformulations and generalizations section of the Wikipedia entry on the Rank and Nullity theorem. –  Adrián Barquero Aug 5 '12 at 19:38
@AdriánBarquero Thank you very much! I do remember the rank nullity theorem! –  Rudy the Reindeer Aug 5 '12 at 19:47
@Clark: I'm looking for an equality of the form $\pm \dim V_1 \pm \dim V_2 \pm \dim V_3 = 0$, with the precise signs to be found by you. I claim that understanding this is necessary and sufficient to answer your question! –  Pete L. Clark Aug 5 '12 at 19:52

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