In "Differential equations driven by rough paths" (Terry Lyons, et al) section 1.4.2 it's claimed that the symmetric part of the tensor:


$\int\_{0 \le u\_1 \le \cdots \le u\_j \le t} \mathrm{d}X\_{u\_1} \otimes \cdots \otimes \mathrm{d}X\_{u\_j}$


is exactly $\frac{1}{j!}(X\_t - X\_0)^{\otimes j}$. It is assumed that $X:[0,T] \to V$ is a Lipschitz path in some finite dimensional vector space $V$.


Is there a simple way to establish this fact? Also I would be grateful for any reference on this type of integrals involving tensors since I've never encountered them before.


**Further background**


Integration with respect to $X$ is defined for any continuous $f: [0,T] \to L(V,W)$ ($W$ a finite dimensional vector space and $L(V,W)$ the space of linear maps from $V$ to $W$) as:


$\int f\_s \mathrm{d}X\_s = \lim \sum\_{i = 0}^{r -1} f\_{t\_i}(X\_{t\_{i+1}} - X\_{t\_i})$


where the limit is taken over partitions $t\_0 = 0 \le t\_1 \le \cdots \le t\_r = T$ whose largest interval is decreasing to $0$ in length. The result belongs to the space $W$.


Hence, to make the above iterated integral fit into the definition (and give a result in $V^{\otimes j}$) it seems to me that one must consider each element $w\_1 \otimes \cdots \otimes w\_i \in V^{\otimes i}$ as an element of $L(V, V^{\otimes i+1})$ via the map $v \mapsto w\_1 \otimes \cdots \otimes w\_i \otimes v$. However after doing this I still have trouble establishing the identity.


As an example, with the above interpretation I get:


$\int\_{0 \le s \le t \le T} \mathrm{d}X\_s \otimes \mathrm{d}X\_t = \frac{1}{2}T^2 v\_1\otimes v\_1 + \frac{2}{3}T^3 v\_1\otimes v\_2 + \frac{1}{3}T^3 v\_2 \otimes v\_1 + \frac{1}{2}T^4 v\_2 \otimes v\_2$


for $X\_t = t v\_1 + t^2 v\_2$ in a $2$-dimensional vector space with basis $(v\_1,v\_2)$. This example shows that the result of the integral itself may be non-symmetric.



An iterated tensor product integral
Let's just view the function $\dot{X}(t) = \frac{d}{dt} X\_t$ as a bounded function taking values in the vector space $V$. The notation $dX\_{u\_1}$ means $\dot{X}(u\_1) du\_1$ Then $dX\_{u\_1} \otimes \cdots \otimes dX\_{u\_k} = \dot{X}(u\_1) \otimes \cdots \otimes \dot{X}(u\_k) du\_1 \cdots du\_k$, should be regarded as a bounded function (or measure) on $[0,t]^k$ taking values in the vector space $V^{\otimes k}$.


Because the operation of projecting to a symmetric part is a linear operation from $V^{\otimes k}$ to itself, you can take it inside of the integral. Let's call this symmetric part $dX\_{u\_1} \cdots dX\_{u\_k}$. I'll use the symbol $u \cdot v$ to denote the symmetric product of of $u, v \in V$ -- that is, $u \cdot u \cdot u = u \otimes u \otimes u$, and the general product is defined by the polarization identity. For example $u \cdot v = ( u \otimes v + v \otimes u) / 2$. The resulting multiplication is commutative.


Therefore your integral is equal to


$\int\_{0 \leq u\_1 \leq \ldots \leq u\_k \leq t} dX\_{u\_1} \cdots dX\_{u\_k}$


Because the symmetric product is commutative, for any permutation $\sigma$ of $\{ 1 \ldots k\}$ you get the same value by integrating any of the permuted regions


$ \int\_{0 \leq u\_1 \leq \ldots \leq u\_k \leq t} dX\_{u\_1} \cdots dX\_{u\_k} = \int\_{0 \leq u\_{\sigma(1)} \leq \ldots \leq u\_{\sigma(k)} \leq t} dX\_{u\_1} \cdots dX\_{u\_k}$


Observe also that the region $0 \leq u\_1 \leq \ldots \leq u\_k \leq t$ is a fundamental domain for the action of $S\_k$ on the cube $[0,t]$ -- that is, you can get the whole cube by permuting the variables $u\_i$, and none of these regions overlap. Therefore, since there are $k!$ such permutations, we sum over permutations to conclude


$ k! \int\_{0 \leq u\_1 \leq \ldots \leq u\_k \leq t} dX\_{u\_1} \cdots dX\_{u\_k} = \int\_{[0,t]^k} dX\_{u\_1} \cdots dX\_{u\_k}$


But, by Fubini and the multilinearity of the symmetric product, the integral on the right is just $(X\_t - X\_0)^k$.


I should remark that this same proof in an even simpler setting also produces the $\frac{1}{k!}$ that appears when you prove the Taylor expansion through iterative use of the fundamental theorem of calculus.


