I need a bit of help understanding the tensor algebra. I don't usually study algebra at all but I've come across it in the form of tensor algebras while learning about the signature of a path. I'm sorry if my questions seem very stupid, I'm just very lost. I am only dealing with the vector space $V = \mathbb{R}^d$ which should hopefully simplify things. I have the tensor algebra defined as follows:
Let $V$ be a real vector space. We define the tensor algebra on $V$, denoted $T(V)$, as the direct sum of $V^{\otimes n}$ for $n = 1,2,\cdots$:
$$
T(V) = \bigoplus_{n=0}^{\infty}V^{\otimes n} = \mathbb{R} \oplus V \oplus V^{\otimes 2} \oplus V^{\otimes 3} \cdots.
$$
and the truncated tensor algebra as:
$$
T(V) =\bigoplus_{n=0}^{N}V^{\otimes n} = \mathbb{R} \oplus V \oplus V^{\otimes 2} \oplus V^{\otimes 3} \cdots V^{\otimes N}.
$$
I feel as though I understand this bit fine. What I am confused about now is to do with notation. In my notes, the author writes: if $\epsilon\in T^N(V)$, then $\xi= (\xi_0,\xi_1,\cdots, \xi_N)$, where $\xi_i \in V^{\otimes i}$ for $i \in \{0,1,\cdots,N\}$.
I thought that an element of $T^N(V)$ was the direct sum of elements from $V^{\otimes i}$ where $i \in \{1,2,\cdots,N\}$. However, the notes make it seem like an element of $T^N(V)$ is a collection of terms each from $V^{\otimes i}$. Should I maybe interpret the element $\xi \in T^N(V)$ as
$$
\xi = (\xi_1, \cdots, \xi_N) = \xi_1 \oplus \cdots \oplus \xi_N?
$$
Now assuming this is the case, my next question is about the multiplication of elements of the tensor algebra. I read that the multiplication of elements in the tensor algebra is actually just the tensor product. So for $A,B \in T^N(V)$, we multiply via $A \otimes B$. I would just like to clarify what this means exactly. Is the following correct:
Suppose $A = (A_0,A_1,A_2)$ and $B = (B_0, B_1,B_2)$. Thus $A_0,B_0 \in V^{\otimes 0}$, $A_1,B_1 \in V^{\otimes 1}$ and $A_2,B_2 \in V^{\otimes 2}$. Then explicitly
$$
A \otimes B = (A_0 \otimes B_0) \oplus (A_0 \otimes B_1) \oplus (A_0 \otimes B_2) \oplus (A_1 \otimes B_0) \oplus (A_1 \otimes B_1) \oplus (A_1 \otimes B_2) \oplus (A_2 \otimes B_0 )\oplus (A_2 \otimes B_1 ) \oplus (A_2 \otimes B_2)
$$
Putting this in the same format as in the notes, we get:
$$
\Bigl((A_0 \otimes B_0), (A_0 \otimes B_1)\oplus (A_1 \otimes B_0),(A_0 \otimes B_2)\oplus (A_1 \otimes B_1)\oplus (A_2 \otimes B_0 ),(A_1 \otimes B_2)\oplus (A_2 \otimes B_1 ), (A_2 \otimes B_2)\Bigr)
$$
Then clearly $A,B \in T^2(V)$ but $A \otimes B \notin T^2(V)$. Is this correct?
Now I'm asking these questions because the definition of the signature of a path $X:[0,T] \rightarrow \mathbb{R}^d$ is as follows:
$$
S(X) = \Bigl(1, X^{1}, X^{2}, \cdots X^{n}, \cdots \Bigr) \in T(V).
$$
So is the signature a sum of all of these $X^i$'s or actually just a collection of them?
Moreover given two paths, elements in the truncated tensor algebra, $\textbf{X} = (X^0,\cdots, X^N)\in T^N(V) $ and $\textbf{Y} = (Y^0, \cdots, Y^N) \in T^N(V)$, I have seen a two versions of a theorem written differently. The first uses the tensor product $\textbf{X} \otimes \textbf{Y}$ and the second uses a product denoted by $*$ defined as:
$$
(\textbf{X}*\textbf{Y})^n = \sum_{j=0}^n X^j \otimes X^{n-j}
$$
Depending on the definition of $\otimes$ these may or may not be the same. If my definition of the tensor product above is correct, then these two products are not equal. However, if I take my definition for the tensor product and remove all the elements who's order is larger than $N$ (by this I mean get rid of the terms in the product that live in $V^{\otimes (N+1)}$, $V^{\otimes (N+2)} \cdots$), then the two are equal (also provided that my understanding of the notation is correct).
Thank you very much in advance – I appreciate answers to any of the above questions!
Best Answer
The first question was already answered in a comment. It is true that if $A, B \in T^2(V)$, then $A \otimes B$ is in general not in $T^2(V)$. The interpretation of $S(X)$ is as in the answer in the comment by Osama Ghani. Note that $T^N(V)$ is not closed with respect to tensor product, but it is closed with respect to $*$. So yes, the two products are different.