# Infine Direct Sum of Vector spaces

1. Aug 5, 2010

### Goldbeetle

Dear all,
I'm reading the tensor part of "A course in modern mathematical physics" by Szekeres and I have trouble understanding a concept that you can find in the attached image of the book page. What are the elements of F(V)? If my understanding of (external) direct sums of vector spaces is correct, the elements of F(V) are arrays of an infinite number components, where each element j of the array is an element of the tensor product of j copies of V. Then, how is the subsequent formal finite formal sum for the typical member of F(V) justified? I'm sure I'm missing some details, perhaps I can't see identifications etc.

Goldbeetle

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2. Aug 5, 2010

### quasar987

True but only a finite number of the components are non zero. So for every element x=(v0, v1, v2,...) of F(V), there is a maximal number r such that vi=0 for all i>r. So it is justified to write x as the formal sum v0+v1+v2+...+vr.

3. Aug 5, 2010

### Goldbeetle

Why only a finite number of components are non zero?

Also, I do not understand the meaning of the "+" sign? Is it a short-hand for the fact that, after identifying for each r each tensor product space V(r) with the subspace of F(V) of the elements (0,0,..., v(r),0,...) , then a generic element of F(V) can be decomposed in a unique way as sum of elements belonging to each V(r)? In an intuitive way, this construction allows to "mix" together possibly wildly different vector spaces in a bigger vector space and give a meaning to a sum of their elements using the "+" sign. We can do algebra with this "+" operation. This works because we cannot "really" (that is not only formally) sum elements belonging to different vector spaces. The "+" is formal when refers to formal addends, whereas it is "real" when it refers to elements of the formal sum that are of the same type, that is, belong to the same subspace of F(V). The "+" sign is overloaded, a C++ programmer would say.

I'm new to this kind of constructions so I get confused at times, I hope that it is at least clear what I do not understand.

Last edited: Aug 5, 2010
4. Aug 5, 2010

### quasar987

Well, this is just by definition of the direct sum:

http://en.wikipedia.org/wiki/Direct_sum_of_modules#Construction_for_an_arbitrary_family_of_modules

Yes, I suppose you could see it that way. But I think it is mostly just a notation that allows you to write an element of the direct sum succintly. Because formally, the direct sum of a familly M_i of R-modules

$$\bigoplus_{i\in I}M_i$$

is defined quite abstractly as the set of all functions

$$f:I\rightarrow \bigcup_{i\in I}M_i$$

such that f(i) belongs to M_i, and f(i)=0 for all but finitely many i. So one way amongst other to note such a function is as the "formal sum"

$$\sum_{j=i}^rf(i_j)$$

where i_j are the indices on which f does not vanish.

Of course it gets awkward when some of the M_i are the same, but if not, as in your case, then the notation is unambiguous. And it has the benefit of somewhat capturing the idea that $\bigoplus_{i\in I}M_i$ takes all the elements of the M_i and "unite" them in one big module.

5. Aug 5, 2010

### Goldbeetle

In the attached page of the book, the number of vector spaces "summed" is infinite, so potentially there are elements with infinitely many non zero components.

6. Aug 5, 2010

### quasar987

By definition, no.

If this were the direct product (aka cartesian product $\times$) being used of the V^(i), then you'd be right. Some elements of F(V) would have infinitely many non zero components. By this is the direct sum, which means, by definition, the element of the direct product for which only finitely many components are non zero.

7. Aug 5, 2010

### Goldbeetle

Quasar987,
I've read more carefully the wikipedia article that you linked above and things are starting to click. The definition given in the book is misleading, it refers to the standard definition of direct sum of vector spaces that is given earlier in the book and does not modify it. Also the meaning of formal sums is not made explicit.

8. Aug 5, 2010

Direct sums satisfy the universal property for the coproduct. That implies that if you have a bunch of vector spaces Vi, then their direct sum is a space V which "contains" each Vi as a subspace, and is the smallest such vector space; that is, no proper subspace of V contains each Vi. That means V must be spanned by each Vi, which means it consists of all vectors that are finite linear combinations of vectors in each Vi.

It just so happens that in the category of vector spaces, the direct sum (coproduct) can be regarded as a subspace of the Cartesian product (product).

9. Aug 6, 2010

### Goldbeetle

Adriank, thanks, but it's way too abstract for me.

10. Aug 6, 2010

I'm just saying that the direct sum is the smallest possible vector space containing each of the summands. (...such that the summands intersect only at 0, anyway, but that's not important for the finite sum thing.)

11. Aug 6, 2010

### Goldbeetle

Is there anyway to "define formal" starting from functions on the set of the infinite direct sum of vector spaces and not null in a finite number of elements?

12. Aug 6, 2010

### Goldbeetle

Ops...I think Quasar987 has done it already!

Last edited: Aug 6, 2010