What is direct sum?

1. Jun 2, 2007

jostpuur

Okey, I have some silly problems with simple definitions.

The usual sum, which I know, of two vector spaces is a set which consists of all sums of the vectors. $$A+B=\{a+b|a\in A,\; b\in B\}$$. Is this the same thing as the direct sum?

I think I saw somewhere (I don't remeber where) a definition, that a direct sum of two vector spaces is just a their union, where zeros are identified. Do I remember this wrong? In some places the direct sum is treated as a vector space, and the union doesn't produce a vector space, so I think that is wrong then...

The wikipedia seems to explain, the the direct sum of two vector spaces is merely their cartesian product, but if it is a cartesian product, why is it called a direct sum then?

2. Jun 2, 2007

mathwonk

this is explained in mY free ALGEBRA BOOK ONLINE.

3. Jun 3, 2007

jostpuur

I got lost into that mapping mess. Well, I'll see how I manage to proceed with this, and maybe return later.

4. Jun 3, 2007

daniel_i_l

A direct sum is like a regular sum with the condition that each value in (A+B) has only one unique representation as a sum of one vector from A and one from B.
It's easy to prove that this happens when the intersection of A and B is {0}.

5. Jun 3, 2007

matt grime

The sum you wrote A+B, implicitly assumes that A and B are subspaces of some larger space C. Otherwise the idea of adding a in A to b in B doesn't make sense.

\oplus or direct sum gets round this minor techincal assumption. As a set it is in natural bijection with AxB. But cartesian products don't have vector space structures.

As has been mentioned, if the former case makes sense, two objects will be isomorphic if A and B intersect in just 0.

But you already knew this: R^2 is RxR is two copies of R added together.

Anyway, for finite cardinals, n, the direct product and direct sums of n copies of a vector space are isomorphic. This fails for infinite direct sums and products.

Last edited: Jun 3, 2007
6. Jun 3, 2007

mathwonk

a direct sum of V,W is a gadget such that maps out of it are determiend by choosing arbitrary maps out of both V and W. e.g. since a map out of V is determiend by sending the elements of a basis anywhere you like, V is always the direcdt sum of the lines spanned by a basis.

more generally, if V,W are spaces their direct sum is a space T, together with maps V-->T, W-->T, such that sending a map T-->Z to the two compositions V-->Z and W-->Z, defiens a bijection between maps out of T and pairs of maps out of V,W.

Thus, after a long explanation I haVENT TIME FOR THIS ECOND, VxW togetehr with the two natural inclusions V-->VxW, and W-->VxW is the direct sum,of V,W, whiele the same space VxW, but togetehr with the two projections VxW-->V, VxW-->W, is the direct product (which clkassifies pairs of maps INTO V,W).

so the sum clasifies pairs of maps OUT of V,W, while the product clasifies maps INTO V,W. Sometimes the same space does both.

then the rpoduct "equals" the sum, but with two diferent maps.

exercise: direct sum of two sets is their disjoint union. their product is the cartesian product.

direct sum of Z and Z is free group on two letters but their product is the cartesian p0roduct ZxZ.

direct sum of two, lets see, algebraic varieties, is their disjoint union, and product is their cartesian product (with appropriate structure).

i cant think of any more interesting examples.

i guess if you believe a map of integers can be an integer dividing another then the sum of two integers is their lcm, and their product is their gcd, i think, but it is late.

Last edited: Jun 3, 2007
7. Jun 3, 2007

mathwonk

confusing as it is, do not give up on the idea of maps as almost more important than objects. i.e. the most fundamental question about two objects is when are they the same? i.e. isomorphic? this means you must know how to define maps on objects. a sum is an object made up of other objects whose maps are determined by arbitrary choices of maps out of the other objetcs.

8. Jun 6, 2007

jostpuur

I guess the definition of direct sum of vector spaces is the same as it is for modules (as given in mathwonk's book on page 10 of 845-3.pdf), if you replace the homomorphisms with linear maps. Correct?

So $$\bigoplus_{\alpha\in A}X_\alpha$$ is that subset (and some mappings with it) of $$\prod_{\alpha\in A} X_\alpha$$, in which all members have only finite number of non-zero components? I have difficulty in seeing how the more abstract definition implies finiteness of the non-zero components. I can see that the finiteness assumption is used in the proof of existance of the sum, but couldn't the needed sums of vectors exist in some cases also without this assumption?

9. Jun 6, 2007

matt grime

linear maps are homomorphisms. A vector space is a module for a field.

The direct sum \oplus X_\alpha must contain (copies of) all finite sums of elements in the X_\alpha, and it must be the smallest such space with this property. By defintion. I don't understand your question: if we drop the finiteness assumption then we've created something bigger than we needed to. So it can't be the direct sum.

10. Jun 6, 2007

jostpuur

The definition I found in mathwonk's book is this:

Let $$\{X_\alpha\}$$ be an indexed collection of modules. A direct sum of this family is a module $$X$$ and a collection of homomorphisms $$\sigma_\alpha:X_\alpha\to X$$ with a following mapping property: For every module $$Y$$ the correspondence taking a homomorphism $$f:X\to Y$$ to the family of compositions $$\{f\circ\sigma_\alpha\}$$ is a bijection between $$\textrm{Hom}(X,Y)$$ and the cartesian product $$\prod \textrm{Hom}(X_\alpha, Y)$$.

The finiteness of non-zero components is not obivous in this definition.

11. Jun 6, 2007

matt grime

Read page 11 of (what I presume to be) those same notes. From that definition it is not clear what the direct sum should be at all. The definition there is retrospectively fixed once we know what a direct sum ought to be. If you didn't have finiteness, then you could in theory have the element x in one of the X_alpha added up withitself infinitely many times.

12. Jun 6, 2007

mathwonk

there is a basic principle in my notes that an object is uniquely determined, up to isomorphism, by knowing its maps into everything.

hence if the direct sum with the finiteness condition satisfies the mapping definition above, then no other object can do so.

i guess the thing that forces the finiteness condition is the idea that maps out should be determined by knowing what they do on the summands.

i.e. knowing the maps out on the summands only determines them further on finite sums of elements of those summands.

it is not a trivial matter to determine what objech has the proeprty of a direct sum, and sometimes such objects may not exist.

the point of this definition is noit that it reveals the structure of the direct sum, but that it states the most important property of the direct sum.

i.e. as stated elsewhere, what we want to know about objects is not so much what they "are" as how thy behave/

13. Jun 6, 2007

mathwonk

take for example the direct sum of two groups, even Z plus Z, in the sense of groups. this would again ba a group G with two maps Z-->G, Z-->G, such that a pair of group maps Z-->H dtermines a unique group map G-->H.

Then if you think about it, you will see that G must be the free group on two generators, i,e, the set of all finite words spelled with two letters a,b, (one for each generator of a summand Z).

Again the finiteness of the words is there (partly) because knowing a map on a and b, only determines it on finite combinations of them.

14. Jun 6, 2007

mathwonk

but the point is that a "definition" in the old sense was really a construction. In the new sense a definition is only a characterization, i.e. a statement of a charaterizing property.

Thus in the odl sense once something has been defined, it does exist. in the enw sense, even after something has been defined, it may or may not exist. A construction is a theorem that the object defined does exist.

so in the old way, an object was first defined, i.e. constructed, and then a theorem was proved that gave its proprties. in the new way, an objects properties are its definition, then it still remains to construct something with those proerties.

so one distinguishes now a definition from a construction. in the old way, even after something was defined you still did not know what it was good for. now the definition itself answers that question.

it is also now possible to se analogies between direct sums of different types, like between free groups and vector spaces. this also leads one to ask for a construction of direct sums in settings where they ahve not yet been constructed.

e.g. amny books do not define, and do not construct diret sums of arbitrary groups/ in the old school this absence wouold not have been noticed, but now the stduents is lways inclined toa sk in every setting whetehr or not diredt sums exist.

this opens ones eyes to structures that should be there but that ahve not been pointed out.

look for direct sums of groups e.g. in Dummit and Foote. on p. 160 they give the wrong, i.e. old definition of direct sum, and hence by not telling you what a direct sum really is in general, they can get away with omitting the concept in the case of groups. note that the concept of direct sums of groups does not even appear in the index.

this is a serious omission in a book supposedly for graduate students. oops, i think this omission of sums of groups occurs as well in my book, but at least one has a chance of noticing it, since i have defined general direct sums. of course it is harder to verify, since my book cleverly has no index.

my excuse is that at time of writing i did not myself know much about free groups and amalgamated sums, and it seemed therefore omittable also for the students, since we were focusing on galois theory. but if we were doing topology and fundamental groups of unions of spaces we woukld have needed these sums of groups, as one sees presumably from a book like hatcher, algebraic topology.

Last edited: Jun 6, 2007
15. Jun 6, 2007

mathwonk

to whack a poor dying horse a coupl more times, the general definition applies to all situations, hence it CANNOT contain any description of specific features of any one of them. the general definition states that property which all direct sums of every stripe have in common.

it seems to me that semi direct product of groups is asort of sum.. i.e. isnt it some gadget such that a map out of it is a pair of maps out of the summands whose images commute? or some such.

im sorry im in a hurry. see my notes for exercises on mapping properties of semi direct products.

16. Jun 6, 2007

jostpuur

Yep, I think I know what that means. For example I first encountered merely properties of real numbers as the definition of real numbers, and later found out about the constructive definition with sequences of rational numbers.

My problem wasn't really in the finest details of direct sum. Actually I was trying to study theory of Lie algebras on my own, but I got stuck because I didn't know what direct sum was all about. I think I'm capable of continuing with it now. Your comments were helpful.

17. Jun 6, 2007

mathwonk

very astute!

18. Jun 6, 2007

mathwonk

here is a quick argumengt for finiteness: if the map out of the sum is to b e determiend by what it does to the summands, then the summands have to generate the sum, but what the summands generate is exactly the submodule of the product where the entries have at most a finite number of non zero entries.

19. Jun 15, 2007

Ultraworld

In my opinion the direct sum of A and B (A and B submoduls/groups/vectors spaces) is just the sum A + B with the extra property that the intersection of A and B equals 0 (not the empty set but 0).

20. Jun 18, 2007

mathwonk

you are considering only the question of what the elements are, and not the more important question of what the maps are. this is the old "naive" point of view.