Direct Sum and Direct Product: Understanding the Differences in Vector Spaces

[Travis091]

The definition (taken from Robert Gilmore's: Lie groups, Lie algebras, and some of their applications):

We have two vector spaces $V_1$ and $V_2$ with bases $\{e_i\}$ and $\{f_i\}$. A basis for the direct product space $V_1\otimes V_2$ can be taken as $\{e_i\otimes f_j\}$. So an element w of this space would look like (summation convention):

$$w=A^{ij}e_i\otimes f_j$$

For the direct sum space $V_1\oplus V_2$, we take as basis: $\{e_1,e_2,...;f_1,f_2,...\}$.
\end of stuff from Gilmore

My question:

If we take $V_1$ to be the x-axis, and $V_2$ to be the y-axis, we can say that the tensor product space is the y=x line. Since any element would look like: $w=A\;\;\hat{x}\otimes \hat{y}$ whereas the direct sum space is spanned by $\{\hat{x},\hat{y}\}$, i.e. it consists of the entire $R^2$.

but it seems weird to me that the tensor product space in this example has a smaller dimension than the direct sum space. Obviously I have a misconception, where is it?

Thanks!

 

[micromass]

The definition (taken from Robert Gilmore's: Lie groups, Lie algebras, and some of their applications):

We have two vector spaces $V_1$ and $V_2$ with bases $\{e_i\}$ and $\{f_i\}$. A basis for the direct product space $V_1\otimes V_2$ can be taken as $\{e_i\otimes f_j\}$. So an element w of this space would look like (summation convention):

$$w=A^{ij}e_i\otimes f_j$$

I don't know why you call this a direct product. It's the tensor product, it's very different from the direct product. In fact, the direct sum and direct product (of two factors) coincide in this case.

For the direct sum space $V_1\oplus V_2$, we take as basis: $\{e_1,e_2,...;f_1,f_2,...\}$.
\end of stuff from Gilmore

My question:

If we take $V_1$ to be the x-axis, and $V_2$ to be the y-axis, we can say that the tensor product space is the y=x line.

In what sense is it the y=x line?

Since any element would look like: $w=A\;\;\hat{x}\otimes \hat{y}$

What do the hats mean on $\hat{x}$?

but it seems weird to me that the tensor product space in this example has a smaller dimension than the direct sum space. Obviously I have a misconception, where is it?

Yes, sometimes the tensor product space has a smaller dimension than the direct sum. A lot of times it has a greater dimension however. Why does this seem weird to you? Do you think that a sum should always be smaller than a product? What about $1\cdot 1<1+1$?

 

[Travis091]

I don't know why you call this a direct product. It's the tensor product, it's very different from the direct product. In fact, the direct sum and direct product (of two factors) coincide in this case.

Some physicists use tensor product and direct product synonymously. The definition given above is the direct product according to Gilmore - he defines the tensor product on groups only - see page 28 of the above mentioned reference.

In what sense is it the y=x line?

In the sense that this space is spanned by $\hat{x}\otimes\hat{y}=(1,1)$ which is a vector in $R^2$ that makes a 45 angle with the x-axis.

What do the hats mean on $\hat{x}$?

Unit vectors.

Yes, sometimes the tensor product space has a smaller dimension than the direct sum. A lot of times it has a greater dimension however. Why does this seem weird to you? Do you think that a sum should always be smaller than a product? What about $1\cdot 1<1+1$?

[/QUOTE]
But the direct (or tensor?) product $R\otimes R$ is simply $R^2$. So I must be misunderstanding something. Or maybe it's Gilmore? (highly doubtful

 

[micromass]

Some physicists use tensor product and direct product synonymously. The definition given above is the direct product according to Gilmore - he defines the tensor product on groups only - see page 28 of the above mentioned reference.

That Gilmore book is awful. He defines the tensor product of vector spaces, not of groups. And he really shouldn't use the term direct product.

In the sense that this space is spanned by $\hat{x}\otimes\hat{y}=(1,1)$ which is a vector in $R^2$ that makes a 45 angle with the x-axis.

I don't follow at all. Why would $\hat{x}\otimes \hat{y} = (1,1)$? That makes no sense at all.

But the direct (or tensor?) product $R\otimes R$ is simply $R^2$. So I must be misunderstanding something. Or maybe it's Gilmore? (highly doubtful

Not at all. We have that $\mathbb{R}\otimes\mathbb{R}$ is the same as $\mathbb{R}$. A basis for $\mathbb{R}$ is simply $\{1\}$ and a basis for $\mathbb{R}\otimes\mathbb{R}$ is $\{1\otimes 1\}$.
In general, if $V$ has dimension $n$ and if $W$ has dimension $m$, then $V\otimes W$ has dimension $m\cdot n$ and $V\oplus W$ has dimension $m+n$.

 

[Travis091]

I don't follow at all. Why would $\hat{x}\otimes \hat{y} = (1,1)$? That makes no sense at all.

So what is this item $\hat{x}\otimes \hat{y}$?

More importantly, what is the vector space spanned by it? I know it is $V_1\otimes V_2$, but what is the geometric picture of this space?

 

[micromass]

So what is this item $\hat{x}\otimes \hat{y}$?

More importantly, what is the vector space spanned by it? I know it is $V_1\otimes V_2$, but what is the geometric picture of this space?

I don't really think there is an easy geometric picture of this.

For $\mathbb{R}\otimes\mathbb{R}$ this is isomorphic to $\mathbb{R}$ and the isomorphism sends $x\otimes y$ to $xy$. So in this sense, the tensor product generalizes the usual product on $\mathbb{R}$.

So that is one way of seeing the tensor product. It provides a multiplication operation on the vector spaces. But unlike the usual multiplication, the multiplication of a vector in $V_1$ and $V_2$ lies in an entirely new space $V_1\otimes V_2$. Still, this is a useful intuition, particularly when you go reading about Fock spaces and such.

The best way of seeing the tensor product is with bilinear maps (or multilinear maps). In that sense, a tensor $x\otimes y$ is in fact related to a bilinear map. For more information, see the free book linear algebra done wrong http://www.math.brown.edu/~treil/papers/LADW/LADW.html Chapter 8.

 

[Travis091]

Thank you for the clarification, and for the excellent reference.