MHB Definition of Direct Product Using UMP - Including Infinite Case

[Math Amateur]

I am reading Chapter 2: Vector Spaces over $$\mathbb{Q}, \mathbb{R} \text{ and } \mathbb{C}$$ of Anthony W. Knapp's book, Basic Algebra.

I need some help with some issues regarding the general UMP-based definition of external and internal direct products ... ...

On page 63, Knapp defines a direct product in terms of a UMP as follows:'View attachment 2944The above it seems ( ... is it? ... ) the definition of the external direct product because Knapp then writes:

View attachment 2945

[ ... Question 1 - is this indeed the definition (only) of the external direct product - which must be then modified/constrained for the internal direct product? ... ]

Knapp then defines the internal direct product as follows:https://www.physicsforums.com/attachments/2946
https://www.physicsforums.com/attachments/2947

BUT ... (Question 2) ... what is the meaning of and nature of $$p_\alpha |_{V_\alpha}$$?

Further, (still Question 2) what does Knapp mean exactly when he says "the restriction $$p_\alpha |_{V_\alpha}$$ is the identity map on $$V_\alpha$$ ... - surely $$p_\alpha |_{V_\alpha}$$ is the identity map on $$V_\alpha$$ anyway ... so how is this an extra condition ... I can only assume that $$p_\alpha |_{V_\alpha}$$ is not necessarily the identity map on $$V_\alpha$$! ... ... but I cannot see how it can be other ... ...?

Can someone please help ...?To indicate the nature of my confusion I will attempt an (very simple) illustrative example ...Let $$A = \{ 1,2,3 \} $$

and suppose that $$V_1, V_2$$, and $$V_3$$ are vector spaces ... ...

Then $$ \prod_{\alpha \in A} V_\alpha = V_1 \times V_2 \times V_3 $$

Let points/elements belonging to $$\prod_{\alpha \in A} V_\alpha$$ be represented by $$\{ v_\alpha \}_{\alpha \in A}$$ so that particular given points might be such points as

$$ \{ v_{1_1} , v_{2_1} , v_{3_1} \} , \{ v_{1_2} , v_{2_2} , v_{3_2} \} , \{ v_{1_6} , v_{2_6} , v_{3_6} \} , \{ v_{1_{23}} , v_{2_{23}} , v_{3_{23}} \}$$, and so on ...

Now consider $$p_1 ( \{ v_\alpha \}_{\alpha \in A} ) $$

We have that:

$$p_1 ( \{ v_{1_1} , v_{2_1} , v_{3_1} \} ) = v_{1_1}
$$

$$p_1 ( \{ v_{1_2} , v_{2_2} , v_{3_2} \} ) = v_{1_2}
$$

$$p_1 ( \{ v_{1_6} , v_{2_6} , v_{3_6} \} ) = v_{1_6}$$

$$p_1 ( \{ v_{1_{23}} , v_{2_{23}} , v_{3_{23}} \} ) = v_{1_{23}}$$and so on ... ...

BUT ... it appears that $$p_1$$ (of course) only operates on $$V_1$$ and is quite naturally the identity map on $$V_1$$ anyway ... whereas I take Knapp to be implying that $$p_1$$ is not necessarily the identity map on $$V_1$$, but needs to be restricted or constrained to be so in the internal direct product ...

Can someone please clarify this issue for me ...

Help will be appreciated ... ...

Peter***NOTE***

In order for MHB members interested in the above post to understand the context of the post, I am including below, the relevant text from Knapp explaining the generalised case (including infinitely many vector spaces) for the direct product for vector spaces ...
View attachment 2948
View attachment 2949
View attachment 2950

 

[Euge]

I am reading Chapter 2: Vector Spaces over $$\mathbb{Q}, \mathbb{R} \text{ and } \mathbb{C}$$ of Anthony W. Knapp's book, Basic Algebra.

I need some help with some issues regarding the general UMP-based definition of external and internal direct products ... ...

On page 63, Knapp defines a direct product in terms of a UMP as follows:'View attachment 2944The above it seems ( ... is it? ... ) the definition of the external direct product because Knapp then writes:

View attachment 2945

[ ... Question 1 - is this indeed the definition (only) of the external direct product - which must be then modified/constrained for the internal direct product? ... ]

No, a direct product is more general than an external direct product. External direct products are constructions which satisfy the UMP for direct products. Namely, given a collection $\{V_{\beta}\}_{\beta\in A}$ of vector spaces over a field $\Bbb F$, its external direct product is defined as the set of all $A$-tuples $(v_{\beta})$, where $v_{\beta}\in V_{\beta}$ for each $\beta$. Define $\Bbb F$-linear maps $p_{\alpha} : \prod_{\beta\in A} V_{\beta} \to V_{\alpha}$ by setting $p_{\alpha}((v_{\beta})) = v_{\alpha}$, for all $\alpha\in A$. Let $U$ be any $\Bbb F$-vector space. Let $L_{\alpha} : U \to V_{\alpha}$ be a collection of $\Bbb F$-linear maps. Define a map $L : U \to \prod_{\beta\in A} V_{\beta}$ such that $L(u) = (L_{\beta}(u))$. Then for each $u\in U$,

$\displaystyle p_{\alpha}L(u) = p_{\alpha}((L_{\beta}(u))) = L_{\alpha}(u)$

for all $\alpha\in A$. In other words, $p_{\alpha}L = L_{\alpha}$. The map $L$ is unique. For if $S : U \to \prod_{\alpha\in A} V_{\alpha}$ is another $\Bbb F$-linear map such that $p_{\alpha}S = L_{\alpha}$ for all $\alpha$, then

$\displaystyle S(u) = (p_{\beta}S(u)) = (L_{\beta}(u)) = L(u)$

for all $u\in U$. Hence, by the UMP for direct products, the external direct product $\prod_{\beta\in A} V_{\beta}$ is a product of $\Bbb F$-vector spaces.

If you're familiar with categories, then you see that I have shown the following: For every field $\Bbb F$, direct products exist in the category of $\Bbb F$-vector spaces.

Knapp then defines the internal direct product as follows:https://www.physicsforums.com/attachments/2946
https://www.physicsforums.com/attachments/2947

BUT ... (Question 2) ... what is the meaning of and nature of $$p_\alpha |_{V_\alpha}$$?

Further, (still Question 2) what does Knapp mean exactly when he says "the restriction $$p_\alpha |_{V_\alpha}$$ is the identity map on $$V_\alpha$$ ... - surely $$p_\alpha |_{V_\alpha}$$ is the identity map on $$V_\alpha$$ anyway ... so how is this an extra condition ... I can only assume that $$p_\alpha |_{V_\alpha}$$ is not necessarily the identity map on $$V_\alpha$$! ... ... but I cannot see how it can be other ... ...?

The $p_{\alpha}$ are not projection maps, if that's what you're thinking. They are generally known as restriction maps, but these maps need not be the identity on the target space. We can construct a vector space $V$, a subspace $U$ of $V$, and a linear map $L : V \to U$ such that $L|U$ is not the identity on $U$. For let $V = \Bbb C$, $U = \Bbb R$, $L : V \to U$ the $\Bbb R$-linear map given by $L(a + bi) = 2a$. The restriction $L|U$ is given by multiplication by two, so it is not the identity on $U$.

Edit: I didn't see the last edit you made until now, so some parts I've written may be redundant.

 

[Deveno]

Let me illustrate with a hopefully transparent example.

"A" direct product of two spaces $U$, and $V$ is a THIRD vector space $W$ TOGETHER with two ADDITIONAL linear maps:

$p_1:W \to U$
$p_2: W \to V$

SUCH THAT, for any other vector space $X$ (yes, now we have 4 spaces we are talking about) and ANY OTHER PAIR (yes, now we have 4 linear transformations to keep track of) of linear maps:

$L_1:X \to U$
$L_2:X \to V$

we have a UNIQUE linear map $L:X \to W$ with $p_1\circ L = L_1$ and $p_2 \circ L = L_2$ (that makes 5 linear maps in all).

"THE" external direct product of two spaces $U$ and $V$ is $U \times V$ together with the vector addition:

$(u_1,v_1) + (u_2,v_2) = (u_1+u_2,v_1+v_2)$ and scalar multiplication:

$\alpha\cdot(u_1,v_1) = (\alpha\cdot u_1,\alpha\cdot v_1)$

It can be seen, using $p_1:U \times V \to U$ given by $p_1(u_1,v_1) = u_1$, and $p_2:U \times V \to V$ given by $p_2(u_1,v_1) = v_1$ that "the" external direct product of $U$ and $V$ is "a" direct product of $U$ and $V$. But it's not necessarily the only one.

Suppose we take $U = \{(x,ax) \in \Bbb R^2\}$ and $V = \Bbb R$.

Then $U \oplus V$ is a direct product of $U$ and $V$. This is a subspace of $\Bbb R^3$.

I claim $\Bbb R^2$ is ALSO a direct product of $U$ and $V$, let:

$p_1(x,y) = (x,ax)$
$p_2(x,y) = y - ax$

Given any linear maps $L_1:W \to U$, and $L_2:W \to V$, we certainly have: $L:W \to \Bbb R^2$ given by:

$L(w) = L_1(w) + (0,L_2(w))$ is a linear map (prove this!).

Now suppose $L_1(w) = (u,au)$ and $L_2(w) = v$.

Then $p_1(L(w)) = p_1((u,au) + (0,v)) = p_1((u,au+v)) = (u,au) = L_1(w)$ ,and:

$p_2(L(w)) = p_2((u,au+v)) = au + v - au = v = L_2(w)$.

Thus $(W,L)$ satisfies the UMP, so we have a direct product. Note these two direct products live "in different dimensions" (one is a 2-dimensional plane in a 2-dimensional space (that is, the entire space), one is a 2-dimensional subspace of a 3-dimensional space).

If we give each of our two target spaces the standard basis, we can write a matrix that defines an isomorphism between them:

$\begin{bmatrix}1&0\\a&0\\0&1 \end{bmatrix}$

So we cannot really say $U \times V = U \oplus V$, they may be two different species. We can say they are isomorphic. In this case, $\Bbb R^2$ is isomorphic to the plane:

$\{(x,y,z) \in \Bbb R^3: ay - x = 0\}$, or in standard form:

$Ax + By + Cz = 0$, the choice: $A = a, B = -1, C = 0$.

**********

Now this is a "special case", we are only dealing with 2 factors. Indeed, we could have gone through all this using injections.

When we have infinitely many factors, it sill makes sense to talk of "an infinite cartesian product" and we can still form the "projection" maps onto each factor space. The problem is when we have infinitely many spaces, the "infinite sum":

$V_1 + V_2 + \cdots$

doesn't make any sense any more. So we can't just "add" the images of the injections all together. In a certain sense, the product is "maximal", and the direct sum is "minimal". What we CAN do, is take all finite sums of the images. This turns out to be "just enough" to give us the vector space structure we need.

The UMP for direct products is the answer to this question:

How can we decompose a vector space into component spaces such that we can factor any map into it from another space in terms of various mappings from the other space into the factors? (We want to find one map from "the other space" and turn it into a bunch of "component maps").

The UMP for direct sums is the answer to this slightly different question:

How can we create a space from other spaces, so that if we have a bunch of maps into a common space from our smaller spaces, we can create ONE map that encodes all of them simultaneously?

 

[Math Amateur]

No, a direct product is more general than an external direct product. External direct products are constructions which satisfy the UMP for direct products. Namely, given a collection $\{V_{\beta}\}_{\beta\in A}$ of vector spaces over a field $\Bbb F$, its external direct product is defined as the set of all $A$-tuples $(v_{\beta})$, where $v_{\beta}\in V_{\beta}$ for each $\beta$. Define $\Bbb F$-linear maps $p_{\alpha} : \prod_{\beta\in A} V_{\beta} \to V_{\alpha}$ by setting $p_{\alpha}((v_{\beta})) = v_{\alpha}$, for all $\alpha\in A$. Let $U$ be any $\Bbb F$-vector space. Let $L_{\alpha} : U \to V_{\alpha}$ be a collection of $\Bbb F$-linear maps. Define a map $L : U \to \prod_{\beta\in A} V_{\beta}$ such that $L(u) = (L_{\beta}(u))$. Then for each $u\in U$,

$\displaystyle p_{\alpha}L(u) = p_{\alpha}((L_{\beta}(u))) = L_{\alpha}(u)$

for all $\alpha\in A$. In other words, $p_{\alpha}L = L_{\alpha}$. The map $L$ is unique. For if $S : U \to \prod_{\alpha\in A} V_{\alpha}$ is another $\Bbb F$-linear map such that $p_{\alpha}S = L_{\alpha}$ for all $\alpha$, then

$\displaystyle S(u) = (p_{\beta}S(u)) = (L_{\beta}(u)) = L(u)$

for all $u\in U$. Hence, by the UMP for direct products, the external direct product $\prod_{\beta\in A} V_{\beta}$ is a product of $\Bbb F$-vector spaces.

If you're familiar with categories, then you see that I have shown the following: For every field $\Bbb F$, direct products exist in the category of $\Bbb F$-vector spaces.

The $p_{\alpha}$ are not projection maps, if that's what you're thinking. They are generally known as restriction maps, but these maps need not be the identity on the target space. We can construct a vector space $V$, a subspace $U$ of $V$, and a linear map $L : V \to U$ such that $L|U$ is not the identity on $U$. For let $V = \Bbb C$, $U = \Bbb R$, $L : V \to U$ the $\Bbb R$-linear map given by $L(a + bi) = 2a$. The restriction $L|U$ is given by multiplication by two, so it is not the identity on $U$.

Edit: I didn't see the last edit you made until now, so some parts I've written may be redundant.

Just got to your post now ... thanks for your help ...

The first thing I note is this ... you write:

"... The $p_{\alpha}$ are not projection maps, if that's what you're thinking. ..."

Yes, I was thinking that ...

... but ? ... you write ..." ... ... Define $\Bbb F$-linear maps $p_{\alpha} : \prod_{\beta\in A} V_{\beta} \to V_{\alpha}$ by setting $p_{\alpha}((v_{\beta})) = v_{\alpha}$, for all $\alpha\in A$. ..."Surely when you write $p_{\alpha}((v_{\beta})) = v_{\alpha}$, for all $\alpha\in A$ ... this is the definition of a projection as it seems to say:

$$p_{\alpha}((v_{\beta})) = p_{\alpha} \{ v_1, v_2, ... \ ... , v_\alpha, ... \ ... \} = v_\alpha $$

which seems to be the definition of a projection ... ... ?

Can you clarify?

Peter

 

[Euge]

Just got to your post now ... thanks for your help ...

The first thing I note is this ... you write:

"... The $p_{\alpha}$ are not projection maps, if that's what you're thinking. ..."

Yes, I was thinking that ...

... but ? ... you write ..." ... ... Define $\Bbb F$-linear maps $p_{\alpha} : \prod_{\beta\in A} V_{\beta} \to V_{\alpha}$ by setting $p_{\alpha}((v_{\beta})) = v_{\alpha}$, for all $\alpha\in A$. ..."Surely when you write $p_{\alpha}((v_{\beta})) = v_{\alpha}$, for all $\alpha\in A$ ... this is the definition of a projection as it seems to say:

$$p_{\alpha}((v_{\beta})) = p_{\alpha} \{ v_1, v_2, ... \ ... , v_\alpha, ... \ ... \} = v_\alpha $$

which seems to be the definition of a projection ... ... ?

Can you clarify?

Peter

Hi Peter,

When I said that the $p_{\alpha}$ are not projection maps, I meant to say that the $p_{\alpha}$ are not projection maps in general. For the construction of the external product, the maps $p_{\alpha}$ are chosen to be natural projections.

 

[Math Amateur]

Hi Peter,

When I said that the $p_{\alpha}$ are not projection maps, I meant to say that the $p_{\alpha}$ are not projection maps in general. For the construction of the external product, the maps $p_{\alpha}$ are chosen to be natural projections.

Hi Euge,

Thanks for that clarification ... BUT ... so sorry to be slow about this, but I need to understand ...

When in Knapp's construction of the direct products (including external and internal) are the $p_{\alpha}$ natural projections and when are they just linear maps ...

For example, are the $p_{\alpha}$ natural projections for (a) the direct product (b) the external direct product (c) the internal direct product ...

By the way, I am assuming we are dealing with three constructions (a), (b) and (c) which have differences from each other ..

Hope you can help,

Peter

 

[Math Amateur]

No, a direct product is more general than an external direct product. External direct products are constructions which satisfy the UMP for direct products. Namely, given a collection $\{V_{\beta}\}_{\beta\in A}$ of vector spaces over a field $\Bbb F$, its external direct product is defined as the set of all $A$-tuples $(v_{\beta})$, where $v_{\beta}\in V_{\beta}$ for each $\beta$. Define $\Bbb F$-linear maps $p_{\alpha} : \prod_{\beta\in A} V_{\beta} \to V_{\alpha}$ by setting $p_{\alpha}((v_{\beta})) = v_{\alpha}$, for all $\alpha\in A$. Let $U$ be any $\Bbb F$-vector space. Let $L_{\alpha} : U \to V_{\alpha}$ be a collection of $\Bbb F$-linear maps. Define a map $L : U \to \prod_{\beta\in A} V_{\beta}$ such that $L(u) = (L_{\beta}(u))$. Then for each $u\in U$,

$\displaystyle p_{\alpha}L(u) = p_{\alpha}((L_{\beta}(u))) = L_{\alpha}(u)$

for all $\alpha\in A$. In other words, $p_{\alpha}L = L_{\alpha}$. The map $L$ is unique. For if $S : U \to \prod_{\alpha\in A} V_{\alpha}$ is another $\Bbb F$-linear map such that $p_{\alpha}S = L_{\alpha}$ for all $\alpha$, then

$\displaystyle S(u) = (p_{\beta}S(u)) = (L_{\beta}(u)) = L(u)$

for all $u\in U$. Hence, by the UMP for direct products, the external direct product $\prod_{\beta\in A} V_{\beta}$ is a product of $\Bbb F$-vector spaces.

If you're familiar with categories, then you see that I have shown the following: For every field $\Bbb F$, direct products exist in the category of $\Bbb F$-vector spaces.

The $p_{\alpha}$ are not projection maps, if that's what you're thinking. They are generally known as restriction maps, but these maps need not be the identity on the target space. We can construct a vector space $V$, a subspace $U$ of $V$, and a linear map $L : V \to U$ such that $L|U$ is not the identity on $U$. For let $V = \Bbb C$, $U = \Bbb R$, $L : V \to U$ the $\Bbb R$-linear map given by $L(a + bi) = 2a$. The restriction $L|U$ is given by multiplication by two, so it is not the identity on $U$.

Edit: I didn't see the last edit you made until now, so some parts I've written may be redundant.

Hi Euge,

Just another clarification ... ...

You write:

" ... ... No, a direct product is more general than an external direct product. External direct products are constructions which satisfy the UMP for direct products. ... ..."

But Knapp defines a direct product as satisfying the UMP ... so I am struggling to see how the direct product and the external direct product differ ...

Can you help?

Peter

 

[Euge]

Re: Definition of DIrect Product Using UMP - Including Infinite Case

Hi Euge,

Thanks for that clarification ... BUT ... so sorry to be slow about this, but I need to understand ...

When in Knapp's construction of the direct products (including external and internal) are the $p_{\alpha}$ natural projections and when are they just linear maps ...

For example, are the $p_{\alpha}$ natural projections for (a) the direct product (b) the external direct product (c) the internal direct product ...

By the way, I am assuming we are dealing with three constructions (a), (b) and (c) which have differences from each other ..

Hope you can help,

Peter

A projection on a vector space V is a linear mapping $p : V \to V$ such that $p\circ p = p$. Nothing in the definition of an external direct product states that the $p_{\alpha}$ are projections. They are just linear maps. Don't make additional assumptions.

 

[Euge]

Re: Definition of DIrect Product Using UMP - Including Infinite Case

Hi Euge,

Just another clarification ... ...

You write:

" ... ... No, a direct product is more general than an external direct product. External direct products are constructions which satisfy the UMP for direct products. ... ..."

But Knapp defines a direct product as satisfying the UMP ... so I am struggling to see how the direct product and the external direct product differ ...

Can you help?

Peter

I think you're missing that an internal direct product is also a direct product. Knapp does mention this. You know that an external product can be different from an internal product. For example, the space $\Bbb R[x] \oplus \Bbb C$ is an external product of real vector spaces, but it is not an internal product. For $\Bbb C$ is not a subspace of $\Bbb R[x]$ and vice versa.

 

[Math Amateur]

​

A projection on a vector space V is a linear mapping $p : V \to V$ such that $p\circ p = p$. Nothing in the definition of an external direct product states that the $p_{\alpha}$ are projections. They are just linear maps. Don't make additional assumptions.

Most helpful Euge ... Thank you

Peter

 

[Deveno]

Hi Euge,

Just another clarification ... ...

You write:

" ... ... No, a direct product is more general than an external direct product. External direct products are constructions which satisfy the UMP for direct products. ... ..."

But Knapp defines a direct product as satisfying the UMP ... so I am struggling to see how the direct product and the external direct product differ ...

Can you help?

Peter

"Projection" means different things in different contexts. It's an over-loaded word. It "generally" means a surjection of some kind.

A direct product is a certain kind of construction we make with "factors". Direct products are not uniquely defined, but all direct products of the same two factors are isomorphic. The external direct product is a SPECIFIC construction we make from two factors that is ONE way to make a direct product. An internal direct product of two subspaces is another way. As the external direct product and the internal direct product of the embedding of the factor spaces in the external direct product are both direct prodcuts, they are isomorphic, and moreover, the isomorphism is uniquely defined (if we want to preserve the factors in a certain "natural" way).

The thing is, the UMP description, or characterization, of a direct product is "portable": in any category which HAS products (not all of them do) the same idea works: we can create direct products in an analogous way in:

Sets
Groups
Abelian groups
Rings
$R$-modules
Vector Spaces
Topological Spaces
$k$-algebras

and other structures I have not listed here. We just substitute in the proper kind of object, and the appropriate arrows (morphisms).

The product may have a different "name" (for historical reasons, mostly) in any given category.

Things are much cleaner for finite-dimensional vector spaces, than for infinite-dimensional ones. I urge you to thoroughly grasp the finite case, before investigating what happens in the infinite-dimensional one, as some things don't generalize well.

Graphically, it's easy to imagine a direct product as a "double-pointed cone", the direct product is the apex, the factor spaces fan out in a ring around the product, with the $p_{\alpha}$ maps leading down to each $V_{\alpha}$.

Our "any other vector space" is at the nadir of an inverted cone (the second half of our "double-cone") and its family of maps lead upwards to the $V_{\alpha}$ on the ring. The upward arrow:

$\displaystyle L:W \to \prod_{\alpha \in A}V_{\alpha}$

makes all the "slice" diagrams commute:$\begin{array}{l}\prod_{\alpha} V_{\alpha}\\ \uparrow \\W \stackrel{L_{\alpha}}{\longrightarrow}V_{\alpha} \end{array}$

(you'll have to imagine the arrow $p_{\alpha}:\prod_{\alpha} V_{\alpha} \to V_{\alpha}$ as I don't have the patience to import a drawing right now).

 

[Math Amateur]

A projection on a vector space V is a linear mapping $p : V \to V$ such that $p\circ p = p$. Nothing in the definition of an external direct product states that the $p_{\alpha}$ are projections. They are just linear maps. Don't make additional assumptions.

Thanks Euge for your help but I need to clarify the situation regarding the $p_{\alpha}$ and the external direct product ...

On a previous post you write:

" ... ... When I said that the [FONT=MathJax_Math]p[FONT=MathJax_Math]α are not projection maps, I meant to say that the [FONT=MathJax_Math]p[FONT=MathJax_Math]α are not projection maps in general. For the construction of the external product, the maps [FONT=MathJax_Math]p[FONT=MathJax_Math]α are chosen to be natural projections. .. ... "In the post directly above you write:

" ... ... Nothing in the definition of an external direct product states that the $p_{\alpha}$ are projections. They are just linear maps. ... "

Forgive me, I cannot square these two comments ... can you clarify/explain ...

Apologies in advance if I am being obtuse ...

Peter

 

[Euge]

Thanks Euge for your help but I need to clarify the situation regarding the $p_{\alpha}$ and the external direct product ...

On a previous post you write:

" ... ... When I said that the [FONT=MathJax_Math]p[FONT=MathJax_Math]α are not projection maps, I meant to say that the [FONT=MathJax_Math]p[FONT=MathJax_Math]α are not projection maps in general. For the construction of the external product, the maps [FONT=MathJax_Math]p[FONT=MathJax_Math]α are chosen to be natural projections. .. ... "In the post directly above you write:

" ... ... Nothing in the definition of an external direct product states that the $p_{\alpha}$ are projections. They are just linear maps. ... "

Forgive me, I cannot square these two comments ... can you clarify/explain ...

Apologies in advance if I am being obtuse ...

Peter

Hi Peter,

You're still confusing the generality of a direct product with the specificity of the external direct product. For example, a tensor product is a product, but is not external. To show that an external direct product is a direct product, we have to specify what the maps $p_{\alpha}$ are. But don't think Knapp is saying that external direct products are equivalent to direct products. You're adding assumptions to the definitions when you should take them as is.

 

[Math Amateur]

Let me illustrate with a hopefully transparent example.

"A" direct product of two spaces $U$, and $V$ is a THIRD vector space $W$ TOGETHER with two ADDITIONAL linear maps:

$p_1:W \to U$
$p_2: W \to V$

SUCH THAT, for any other vector space $X$ (yes, now we have 4 spaces we are talking about) and ANY OTHER PAIR (yes, now we have 4 linear transformations to keep track of) of linear maps:

$L_1:X \to U$
$L_2:X \to V$

we have a UNIQUE linear map $L:X \to W$ with $p_1\circ L = L_1$ and $p_2 \circ L = L_2$ (that makes 5 linear maps in all).

"THE" external direct product of two spaces $U$ and $V$ is $U \times V$ together with the vector addition:

$(u_1,v_1) + (u_2,v_2) = (u_1+u_2,v_1+v_2)$ and scalar multiplication:

$\alpha\cdot(u_1,v_1) = (\alpha\cdot u_1,\alpha\cdot v_1)$

It can be seen, using $p_1:U \times V \to U$ given by $p_1(u_1,v_1) = u_1$, and $p_2:U \times V \to V$ given by $p_2(u_1,v_1) = v_1$ that "the" external direct product of $U$ and $V$ is "a" direct product of $U$ and $V$. But it's not necessarily the only one.

Suppose we take $U = \{(x,ax) \in \Bbb R^2\}$ and $V = \Bbb R$.

Then $U \oplus V$ is a direct product of $U$ and $V$. This is a subspace of $\Bbb R^3$.

I claim $\Bbb R^2$ is ALSO a direct product of $U$ and $V$, let:

$p_1(x,y) = (x,ax)$
$p_2(x,y) = y - ax$

Given any linear maps $L_1:W \to U$, and $L_2:W \to V$, we certainly have: $L:W \to \Bbb R^2$ given by:

$L(w) = L_1(w) + (0,L_2(w))$ is a linear map (prove this!).

Now suppose $L_1(w) = (u,au)$ and $L_2(w) = v$.

Then $p_1(L(w)) = p_1((u,au) + (0,v)) = p_1((u,au+v)) = (u,au) = L_1(w)$ ,and:

$p_2(L(w)) = p_2((u,au+v)) = au + v - au = v = L_2(w)$.

Thus $(W,L)$ satisfies the UMP, so we have a direct product. Note these two direct products live "in different dimensions" (one is a 2-dimensional plane in a 2-dimensional space (that is, the entire space), one is a 2-dimensional subspace of a 3-dimensional space).

If we give each of our two target spaces the standard basis, we can write a matrix that defines an isomorphism between them:

$\begin{bmatrix}1&0\\a&0\\0&1 \end{bmatrix}$

So we cannot really say $U \times V = U \oplus V$, they may be two different species. We can say they are isomorphic. In this case, $\Bbb R^2$ is isomorphic to the plane:

$\{(x,y,z) \in \Bbb R^3: ay - x = 0\}$, or in standard form:

$Ax + By + Cz = 0$, the choice: $A = a, B = -1, C = 0$.

**********

Now this is a "special case", we are only dealing with 2 factors. Indeed, we could have gone through all this using injections.

When we have infinitely many factors, it sill makes sense to talk of "an infinite cartesian product" and we can still form the "projection" maps onto each factor space. The problem is when we have infinitely many spaces, the "infinite sum":

$V_1 + V_2 + \cdots$

doesn't make any sense any more. So we can't just "add" the images of the injections all together. In a certain sense, the product is "maximal", and the direct sum is "minimal". What we CAN do, is take all finite sums of the images. This turns out to be "just enough" to give us the vector space structure we need.

The UMP for direct products is the answer to this question:

How can we decompose a vector space into component spaces such that we can factor any map into it from another space in terms of various mappings from the other space into the factors? (We want to find one map from "the other space" and turn it into a bunch of "component maps").

The UMP for direct sums is the answer to this slightly different question:

How can we create a space from other spaces, so that if we have a bunch of maps into a common space from our smaller spaces, we can create ONE map that encodes all of them simultaneously?

Thanks Deveno ... this post is so revealing and helpful regarding the theory of direct products ... it starts to answer one of my problems ... how do a set of vector spaces lead to more than one direct product ...

Just some points ... in your post you write ... ...

"... ... "THE" external direct product of two spaces $U$ and $V$ is $U \times V$ together with the vector addition:

$(u_1,v_1) + (u_2,v_2) = (u_1+u_2,v_1+v_2)$ and scalar multiplication:

$\alpha\cdot(u_1,v_1) = (\alpha\cdot u_1,\alpha\cdot v_1)$ ... ... "

Well! That was a major clarification for a start ... because when Knapp introduced the section on direct sums and products he writes ...

View attachment 2953

In the above, when he defines direct product, I am taking it that he is defining EXTERNAL direct product ... Is that right?Now in your post you state (and demonstrate) that $$W$$ is external direct product of $$U$$ and $$V$$.

That is, symbolically, $$W = U \times V$$ together with the componentwise addition and the scalar multiplication.

(in doing the above, you use $$p_1$$ and $$P_2$$ as $p_1(u_1,v_1) = u_1$ and $p_2(u_1,v_1) = v_1$ - that is projections ... can you confirm that $$p_1, p_2$$ are always classical projections of this form for the external direct product?You then give concrete form to U and V when you write:

$U = \{(x,ax) \in \Bbb R^2\}$ and $V = \Bbb R$.

So $$U \times V = \{ ( ( x, ax ) , r ) \ | \ x, r \in \mathbb{R} , a \in \mathbb{F} \}$$

Can you show/explain exactly how this is a subset of $$\mathbb{R}^3 $$

(It seems intuitively so ... yet this example is so important to an understanding of direct products I wish to fully understand the situation ...)

I am being very careful to fully understand your example because it is so fundamental and interesting ... I have read a number of texts that prove isomorphisms between direct products of a set of vector spaces, modules or rings, but never indicate or demonstrate how a set of vector spaces, modules or rings might lead to several different direct products ... at last your example seems to show this! ... great! ... thank you ...

Peter

 

[Math Amateur]

Hi Peter,

You're still confusing the generality of a direct product with the specificity of the external direct product. For example, a tensor product is a product, but is not external. To show that an external direct product is a direct product, we have to specify what the maps $p_{\alpha}$ are. But don't think Knapp is saying that external direct products are equivalent to direct products. You're adding assumptions to the definitions when you should take them as is.

Thanks again Euge ... yes, getting the point ... :-) ... slowly though ... still working on this ... working through some other texts such as Bland's book on rings and modules ...

Do you have a text whose coverage on direct sums and products is good ...

Peter

 

[Deveno]

In an inner product space, the projection of the vector $v$ in the direction of $u$ is the vector:

$p_u(v) = \dfrac{\langle u,v\rangle}{\langle u,u\rangle}u$.

In particular, if $V = \Bbb R^2$ and $v = (v_1,v_2)$ with the usual Euclidean inner product ("dot product"), we have:

$p_{e_1}(v) = \dfrac{e_1\cdot v}{e_1\cdot e_1}e_1$

$= (1,0)\cdot(v_1,v_2)(1,0) = ((1)(v_1) + (0)(v_2))(1,0) = (v_1,0)$

and similarly $p_{e_2}(v_1,v_2) = (0,v_2)$.

These mappings have this property:

$p_{e_1} \circ p_{e_1} = p_{e_1}$
$p_{e_2} \circ p_{e_2} = p_{e_2}$

And so, in general, a linear mapping $P: V \to V$ with the property $P^2 = P$ is called a projection.

If $U = \text{im }P$, it is also called a projection of $V$ ONTO $U$. Note that for $u \in U$:

$u = P(v) = P(P(v)) = P(u)$, so that $P|_U$ ($P$ restricted to the domain $U$) is the identity map.

So in the INTERNAL direct product, the "direct product factor maps" (the "$p$'s") are in fact, projections in this sense.

In many categories (groups, sets, rings, etc.) the $p$ maps are, in fact, surjective morphisms, which are likewise called "projections" (or: the "canonical projections") by extension of this idea.

This data (the description of the $p$ maps) is not part of the definition of the UMP, and there are categories with a product construction in which said maps are not "projections" (they aren't even functions), here is an example:

Let $\Bbb N$ be the natural numbers. The objects of our category will be natural numbers (1,2,3,etc.).

For 2 natural numbers, $k,m$, we define there to be a UNIQUE arrow $k \to m$ if and only if $m|k$.

This defines a category:

If we have an arrow $k \to m$, and an arrow $m \to n$, we can form an arrow $k \to n$ as the composite of our two arrows, since divisibility is transitive.

For any object $k$, we have a (unique arrow) $k \to k$, since for all $k$, we have $k|k$.

Now given $k,m$, consider $u = \text{lcm}(k,m)$. We obviously have two maps:

$u \to k$
$u \to m$

since $k|u$ and $m|u$ (by definition of "common multiple").

Now suppose $w$ is any natural number for which $k|w$ and $m|w$ so that there exists arrows:

$w \to k$
$w \to m$

So $w = ak = bm$.

It is a well-know fact from number theory, that in this case, $u|w$, which in our category means there is an arrow:

$w \to u$ (and this arrow is unique, because all arrows in this category are).

Since $k|u,\ m|u$ and $u|w$ the diagrams (the commutative triangles) formed by the arrows "the only way that makes sense" satisfies the UMP property for a direct product.

Hence in the poset $\Bbb N$, partially ordered by divisibility (with the arrows going from "multiple to factor"), the least common multiple IS the product. Let's say we call this category $N$.

Then $N$ as we have defined it here, does NOT possesses a co-product; however, we can form the opposite category $N^{\text{op}}$ (all we do is reverse the way the arrows go) and this DOES have a co-product, the greatest common divisor, which satisfies the same UMP a direct sum does in vector spaces.

In this instance of these constructions, it does not even make sense to call the arrows:

$u \to k$
$u \to m$

"projections" (and we do not NEED to do so, we just have to see the UMP diagram commutes).

I believe that what Euge is trying to say, it that the mappings:

$\displaystyle p_{\alpha}: \prod_{\alpha \in A} V_{\alpha} \to V_{\alpha}$

are not required before-hand to even be surjective, all that is required is that the map:

$\displaystyle L:W \to \prod_{\alpha \in A} V_{\alpha}$ "factor through them", in the sense that:

$p_{\alpha} \circ L = L_{\alpha}$ for each $\alpha \in A$.

******************

It turns out you CAN show that in the case of VECTOR SPACES, these are indeed ONTO maps:

Take $W = V_{\alpha_0}$ for some fixed $\alpha_0 \in A$, and consider the family:

$L_{\alpha}: V_{\alpha_0} \to V_{\alpha}$, where:

$L_{\alpha} = 1_{V_{\alpha_0}}$ if $\alpha = \alpha_0$
$L_{\alpha} = 0$, otherwise.

We have: $p_{\alpha_0} \circ L = 1_{V_{\alpha_0}}$

and it follows that $p_{\alpha_0}$ is surjective, and $L$ is injective (since identity functions are bijective).

******************

To answer your last question, what you are actually doing is asking:

"What is the set-isomorphism (bijection) between $A^3$ and$ (A^2) \times A$?"

Specifically, it is this:

$(a_1,a_2,a_3) \mapsto ((a_1,a_2),a_3)$

Since this IS a bijection, no confusion can arise if we omit the "inner parentheses".

For example, when we embed the $xy$-plane as an "entity-unto-itself" into real 3-space, we just "pad it with 0's":

$(x,y) \mapsto (x,y,0)$

The 0-space is "dimensionless" and does not affect the algebra. This isomorphism is, in fact, the INJECTION map of the direct sum (as opposed to direct product).

Stare at this equation long and hard until it hits you:

$(v_1,v_2,\dots,v_n) = (v_1,0,\dots,0) + (0,v_2,\dots,0) + \cdots + (0,\dots,0,v_n)$

and ask yourself: "does the vector $(0,\dots,v_k,\dots,0)$ really tell me anything except that its $k$-th coordinate is $v_k$?

 

[Euge]

Thanks again Euge ... yes, getting the point ... :-) ... slowly though ... still working on this ... working through some other texts such as Bland's book on rings and modules ...

Do you have a text whose coverage on direct sums and products is good ...

Peter

I have Beachy's book on rings and modules. The topic of direct sums and products is covered in Chapter 2, Section 2.2. The material is a bit terse, but it contains several useful theorems for identifying a module as a direct sum.

 

[Math Amateur]

Let me illustrate with a hopefully transparent example.

"A" direct product of two spaces $U$, and $V$ is a THIRD vector space $W$ TOGETHER with two ADDITIONAL linear maps:

$p_1:W \to U$
$p_2: W \to V$

SUCH THAT, for any other vector space $X$ (yes, now we have 4 spaces we are talking about) and ANY OTHER PAIR (yes, now we have 4 linear transformations to keep track of) of linear maps:

$L_1:X \to U$
$L_2:X \to V$

we have a UNIQUE linear map $L:X \to W$ with $p_1\circ L = L_1$ and $p_2 \circ L = L_2$ (that makes 5 linear maps in all).

"THE" external direct product of two spaces $U$ and $V$ is $U \times V$ together with the vector addition:

$(u_1,v_1) + (u_2,v_2) = (u_1+u_2,v_1+v_2)$ and scalar multiplication:

$\alpha\cdot(u_1,v_1) = (\alpha\cdot u_1,\alpha\cdot v_1)$

It can be seen, using $p_1:U \times V \to U$ given by $p_1(u_1,v_1) = u_1$, and $p_2:U \times V \to V$ given by $p_2(u_1,v_1) = v_1$ that "the" external direct product of $U$ and $V$ is "a" direct product of $U$ and $V$. But it's not necessarily the only one.

Suppose we take $U = \{(x,ax) \in \Bbb R^2\}$ and $V = \Bbb R$.

Then $U \oplus V$ is a direct product of $U$ and $V$. This is a subspace of $\Bbb R^3$.

I claim $\Bbb R^2$ is ALSO a direct product of $U$ and $V$, let:

$p_1(x,y) = (x,ax)$
$p_2(x,y) = y - ax$

Given any linear maps $L_1:W \to U$, and $L_2:W \to V$, we certainly have: $L:W \to \Bbb R^2$ given by:

$L(w) = L_1(w) + (0,L_2(w))$ is a linear map (prove this!).

Now suppose $L_1(w) = (u,au)$ and $L_2(w) = v$.

Then $p_1(L(w)) = p_1((u,au) + (0,v)) = p_1((u,au+v)) = (u,au) = L_1(w)$ ,and:

$p_2(L(w)) = p_2((u,au+v)) = au + v - au = v = L_2(w)$.

Thus $(W,L)$ satisfies the UMP, so we have a direct product. Note these two direct products live "in different dimensions" (one is a 2-dimensional plane in a 2-dimensional space (that is, the entire space), one is a 2-dimensional subspace of a 3-dimensional space).

If we give each of our two target spaces the standard basis, we can write a matrix that defines an isomorphism between them:

$\begin{bmatrix}1&0\\a&0\\0&1 \end{bmatrix}$

So we cannot really say $U \times V = U \oplus V$, they may be two different species. We can say they are isomorphic. In this case, $\Bbb R^2$ is isomorphic to the plane:

$\{(x,y,z) \in \Bbb R^3: ay - x = 0\}$, or in standard form:

$Ax + By + Cz = 0$, the choice: $A = a, B = -1, C = 0$.

**********

Now this is a "special case", we are only dealing with 2 factors. Indeed, we could have gone through all this using injections.

When we have infinitely many factors, it sill makes sense to talk of "an infinite cartesian product" and we can still form the "projection" maps onto each factor space. The problem is when we have infinitely many spaces, the "infinite sum":

$V_1 + V_2 + \cdots$

doesn't make any sense any more. So we can't just "add" the images of the injections all together. In a certain sense, the product is "maximal", and the direct sum is "minimal". What we CAN do, is take all finite sums of the images. This turns out to be "just enough" to give us the vector space structure we need.

The UMP for direct products is the answer to this question:

How can we decompose a vector space into component spaces such that we can factor any map into it from another space in terms of various mappings from the other space into the factors? (We want to find one map from "the other space" and turn it into a bunch of "component maps").

The UMP for direct sums is the answer to this slightly different question:

How can we create a space from other spaces, so that if we have a bunch of maps into a common space from our smaller spaces, we can create ONE map that encodes all of them simultaneously?

Hi Deveno,

Thanks again for the above example ... it is so helpful! ... it should be placed in every abstract algebra text at the advanced undergraduate level and above!

Just a clarification ... you show that the external direct product is indeed a direct product and then you take a specific example as follows:

" ... ... Suppose we take $U = \{(x,ax) \in \Bbb R^2\}$ and $V = \Bbb R$.

Then $U \oplus V$ is a direct product of $U$ and $V$. This is a subspace of $\Bbb R^3$. ... ... "

Question 1:

You seem to be writing the external direct product as $U \oplus V$ - that is direct sum notation ... why are you referring to it this way and not as $$U \times V$$?

Question 2:

You then give a second example which you also show is a direct product which you seem to refer to as $$U \times V$$ since you write:

" ... ... So we cannot really say $U \times V = U \oplus V$, they may be two different species. We can say they are isomorphic. In this case, $\Bbb R^2$ is isomorphic to the plane: etc ... ... "

So you seem to be referring to the previous external direct product as $ U \oplus V$ (direct sum notation) and the new direct sum as $$U \times V$$.

Can you explain what is going on ...

(Sorry to be pedantic about this example ... but it is so key to understanding direct products!)

Hope you can help ... ...

Peter

 

[Deveno]

Hi Deveno,

Thanks again for the above example ... it is so helpful! ... it should be placed in every abstract algebra text at the advanced undergraduate level and above!

Just a clarification ... you show that the external direct product is indeed a direct product and then you take a specific example as follows:

" ... ... Suppose we take $U = \{(x,ax) \in \Bbb R^2\}$ and $V = \Bbb R$.

Then $U \oplus V$ is a direct product of $U$ and $V$. This is a subspace of $\Bbb R^3$. ... ... "

Question 1:

You seem to be writing the external direct product as $U \oplus V$ - that is direct sum notation ... why are you referring to it this way and not as $$U \times V$$?

Question 2:

You then give a second example which you also show is a direct product which you seem to refer to as $$U \times V$$ since you write:

" ... ... So we cannot really say $U \times V = U \oplus V$, they may be two different species. We can say they are isomorphic. In this case, $\Bbb R^2$ is isomorphic to the plane: etc ... ... "

So you seem to be referring to the previous external direct product as $ U \oplus V$ (direct sum notation) and the new direct sum as $$U \times V$$.

Can you explain what is going on ...

(Sorry to be pedantic about this example ... but it is so key to understanding direct products!)

Hope you can help ... ...

Peter

In this example, it really doesn't matter. I wrote $U\oplus V$ just to indicate that $U$ and $V$ could both be considered part of $\Bbb R^3 = \Bbb R^2 \oplus \Bbb R$. In a concrete finite-dimensional setting, you can't really distinguish between a direct sum and a direct product, each of one is the other.

Arguably, I should have defined $U$ as $\{(x,ax,0) \in \Bbb R^3\}$ and similarly, for $V$, I hope you can see this doesn't affect the algebra. Padding with extra 0-coordinates doesn't change our sums at all.

I just wanted to underscore the idea that the $p$ maps need not be the "standard" projections. They just need to be a pair of maps that make the UMP work.

Whether we talk of $(x,ax)\oplus(y)$ or $(x,ax,y)$, or $((x,ax),y)$ is largely a matter of semantics. In effect, when we write:

$\Bbb R\oplus \Bbb R = \Bbb R^2$

we are (concretely) thinking of some BASIS of $\Bbb R^2$. This choice of basis is not UNIQUE, there is nothing that dictates our basis need be $\{(1,0),(0,1)\}$ or even $\{(a,0),(0,b)\}$ (with $a,b \neq 0$). We can create a direct product (sum) of $\Bbb R$ with itself in many, many ways-these ways are often called coordinate systems.

The "standard" projection functions lead us to the "standard" basis. We can identify my $U$ with $\Bbb R$ like so:

$(x,ax) \mapsto x$ (this is a bijection, indeed, an isomorphism of vector spaces)

and my $V$ with $\Bbb R$ similarly: $(0,y) \mapsto y$. If you prefer, think of the embedded injections of $U$ and $V$ into $\Bbb R^3$ as $\tilde{U}$ and $\tilde{V}$. The basis of $\Bbb R^2$ we obtain from $U,V$ is $\{(1,a),(0,1)\}$ (this is a basis for ANY choice of $a$).

If $a = 0$ this is essentially the standard direct product. If $a = 1$ it is not.

The difference between direct sum and direct product only becomes apparent when we have an infinite number of factors. The important thing to realize about a direct sum is not that:

$\displaystyle \bigoplus_{\alpha \in A} V_{\alpha}$

lives in some "bigger space", but rather just that it IS a space, and we have certain "inclusion" maps:

$\displaystyle i_{\alpha}: V_{\alpha} \to \bigoplus_{\alpha \in A} V_{\alpha}$

These maps will be injective maps, but may not (strictly speaking) be inclusions.

******************

You will find that for finite-dimensional vector spaces, many authors write $v_1 + v_2$ instead of $(v_1,v_2)$. and often do not even mention the difference between a direct product, and a direct sum. Books on modules are usually a tad more careful about such things. Partly, this is because modules need not be free, but vector spaces always are, so vector spaces have "nice enough" properties that we can take the fact they always have a basis and exploit it vigorously.