I Understanding Z2 Graded Vector Spaces: Definition and Examples

[Silviu]

Hello! I just started reading about $Z_2$ graded vector spaces (and graded vector spaces in general) and I want to make sure I understand from the beginning. So the definition, as I understand it, is that a graded vector space can be decomposed into subspaces of degree 0 and 1. So $V=V_1 \oplus V_2$. I am confused about this, does this mean that $R^2=R \oplus R$ is a $Z_2$ graded vector space? And $R^3=R^2 \oplus R$, too? And in the case of $R^3$, does it matter the order of $R$ and $R^2$? I see that one can define the parity reversed spaced, in which $V_1$ and $V_2$ are interchanged, so I guess it makes a difference, but I am confused why, as I thought that the direct sum is commutative. Can someone please explain this to me, in not advanced terms as I really know nothing about graded vector spaces, yet (I would really appreciate a clear example, too, in case what I said above is wrong).

 

[fresh_42]

Hello! I just started reading about $Z_2$ graded vector spaces (and graded vector spaces in general) and I want to make sure I understand from the beginning. So the definition, as I understand it, is that a graded vector space can be decomposed into subspaces of degree 0 and 1. So $V=V_1 \oplus V_2$. I am confused about this, does this mean that $R^2=R \oplus R$ is a $Z_2$ graded vector space? And $R^3=R^2 \oplus R$, too? And in the case of $R^3$, does it matter the order of $R$ and $R^2$? I see that one can define the parity reversed spaced, in which $V_1$ and $V_2$ are interchanged, so I guess it makes a difference, but I am confused why, as I thought that the direct sum is commutative. Can someone please explain this to me, in not advanced terms as I really know nothing about graded vector spaces, yet (I would really appreciate a clear example, too, in case what I said above is wrong).

A graded vector space is just a direct sum of vector spaces, any. The direct summands being called homogenous of a certain degree. So there is not much gain in this concept. Things become different if we consider algebras, i.e. vector spaces with a multiplication. In this case it is interesting how multiplication of homogenous elements behave: in case $V^i \cdot V^j \subseteq V^k$ and $i +j = k$ is according to the group structure of $\mathbb{Z}_2$, we speak of a $\mathbb{Z}_2-$graded algebra.

The Graßmann- and tensor algebras over a vector space are an example of a $\mathbb{Z}$ grading, the super algebras in string theory are an example of $\mathbb{Z}_2-$graded (Lie) algebras.

Wikipedia has a good overview with links to certain examples: https://en.wikipedia.org/wiki/Graded_(mathematics)

 

[Silviu]

A graded vector space is just a direct sum of vector spaces, any. The direct summands being called homogenous of a certain degree. So there is not much gain in this concept. Things become different if we consider algebras, i.e. vector spaces with a multiplication. In this case it is interesting how multiplication of homogenous elements behave: in case $V^i \cdot V^j \subseteq V^k$ and $i +j = k$ is according to the group structure of $\mathbb{Z}_2$, we speak of a $\mathbb{Z}_2-$graded algebra.

The Graßmann- and tensor algebras over a vector space are an example of a $\mathbb{Z}$ grading, the super algebras in string theory are an example of $\mathbb{Z}_2-$graded (Lie) algebras.

Wikipedia has a good overview with links to certain examples: https://en.wikipedia.org/wiki/Graded_(mathematics)

Thank you for this! I looked at the article on Wikipedia. My main question is, i guess, as far as I understand any vector space with a countable basis seems to be a graded vector space. So, what is the difference between a vector space with a countable basis and a graded vector space? Also, about the part with defining the parity reversed spaced for $Z_2$ graded vector space, I am not sure I understand why you need that, as the direct sum is commutative. Is it just convenient from a notation point of view (when you define grade reversing transformation for example) or is there anything deeper to it?

 

[fresh_42]

Thank you for this! I looked at the article on Wikipedia. My main question is, i guess, as far as I understand any vector space with a countable basis seems to be a graded vector space. So, what is the difference between a vector space with a countable basis and a graded vector space?

Any decomposition into a direct sum is technically a grading. The question is whether you want to have one of the many, that are possible, be labeled as grading, i.e. what you're going to achieve. As said, usually there is a multiplication involved.

Also, about the part with defining the parity reversed spaced for $Z_2$ graded vector space, I am not sure I understand why you need that, as the direct sum is commutative.

Same as before. As long as you only consider addition, there won't by any gains from the grading. The moment you start multiplying, commutativity isn't any longer guaranteed.

Is it just convenient from a notation point of view (when you define grade reversing transformation for example) or is there anything deeper to it?

Grading is always about the multiplication structure, not the addition. At least I cannot think of any useful application of a graded vector space. At least a sort of operation on it should be necessary in my opinion.