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I am considering a second-order tensor $\mathbf T$ seen as a linear maps on vector spaces, acting on a vector $\mathbf v$ to create a new vector $\mathbf u$, thus, $\mathbf u= \mathbf{T}(\mathbf{v})$. Starting from this very basic fact, how do we properly introduce the notation $\mathbf u= \mathbf{T}(\mathbf{v})=\mathbf{T}\cdot \mathbf{v}$ where the "$\cdot$" denotes the tensor product contracted once (or inner product, or scalar product when we deal with vectors)? In other words, how do we know that these two notations are interchangeable?

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    $\begingroup$ For a linear function of $\mathbb R$, $f(x)=F.x$. I think it is the same thing here: $T$ is a linear mapping over $v$, so $T(v)$ (= image of $v$ by a function) is equal to $T.v$. The notations are interchangeable because by definition of the function $T$ as linear, $T$ can be written $T.v$. Does it help? $\endgroup$ – anderstood Oct 12 '15 at 21:49
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If you are asking about notation, the comment of @anderstood is spot on. If $f:\mathbb{R}\rightarrow\mathbb{R}$ is any real function, writing $fx$ instead of $f(x)$ is confusing, because $fx$ could be the product of $f$ and $x$. However, if $f$ is a linear function (linear in a linear algebraic sense, eg. homogenous linear), then $f(x)=ax$, so identifying $f$ with $a$ gives us $f(x)=fx$.

The same notion is carried over into linear algebra, except in that case, $Tx$ is even less confusing, because there is no product operation on a general vector space and $T$ is NOT an element of the vector space anyways, so $Tx=T(x)$ is the only possible interpretation, even for nonlinear maps (unless you are explicitly identifying everything with matrices...).

So if we declare $T(x)\equiv T\cdot x$, this is a perfectly fine and unambigous thing to do.

If you are asking to prove if the action of $T$ on $x$ is the same as the contraction performed on $T\otimes x$ over the last two arguments, we can invoke the universal factorization property of tensor products, that essentially states that any multilinear function over the direct sum of some vector spaces has a unique representation as a linear map over the corresponding tensor spaces.

Now, if $T$ is seen as a linear operator on some finite dimensional vector space $V$ (over $\mathbb{F}$), then the space $\hom(V)$ of such maps is naturally isomorphic to the tensor product $V\otimes V^\ast$, and the action $T(x)$ can be seen as a bilinear map $(\cdot,\cdot):V\otimes V^\ast\times V\rightarrow V$. By the universal factorization property, there is a unique linear map, which we now call $\mathrm{tr}^2_1:V\otimes V^\ast\otimes V\rightarrow V$ so that $(T,x)=T(x)=\mathrm{tr}^2_1(T\otimes x)$, which is what we call "contraction over the first lower index and second upper index".

By this argument, any tensor action can be seen as a contraction performed over a tensor product.

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  • $\begingroup$ Well, yes interesting answer but a bit too advanced for my needs. When you say "So if we declare T(x)≡T⋅x, this is a perfectly fine and unambigous thing to do.", you seem to go a bit fast because "$\cdot$" has a very specific meaning, and it might not be that clear that all possible actions of T on x can be completely expressed as an "inner product" T⋅x. $\endgroup$ – pluton Oct 12 '15 at 23:43
  • $\begingroup$ @pluton I'm sorry, I don't understand you, $T\cdot x$ has no specific meaning, unless the vector space in question is $\mathbb{R}^n$, and it is explicitly identified with the matrix space $\mathbb{R}^{n\times 1}$, with all linear transforms identified with $\mathbb{R}^{n\times n}$. If that is not the case, there is no intrinsic meaning to $T\cdot x$, we can however define it as $T\cdot x=T(x)$. $\endgroup$ – Bence Racskó Oct 13 '15 at 10:21
  • $\begingroup$ ah ok: just because in my question, I said it was the usual inner product and I was not sure whether you kept this meaning or not in your answer. $\endgroup$ – pluton Oct 13 '15 at 12:31
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Given a $(1,1)$ tensor as a map $$ \tau\colon V\times V^*\to\mathbb{C} $$ such element can be always expressed as $\tau = \tau^{\mu}_{\phantom{\mu}\nu}e_{\mu}\otimes \alpha^{\nu}$, with $(e_{\mu}, \alpha^{\nu})$ being a basis of $V, V^*$, respectively.

Using your notation, let us act with $\tau$ upon a vector $v\in V$. We have $$ \tau(v) = \left(\tau^{\mu}_{\phantom{\mu}\nu}e_{\mu}\otimes \alpha^{\nu}\right)(v)= \left(\tau^{\mu}_{\phantom{\mu}\nu}e_{\mu}\right)\cdot\alpha^{\nu}(v) = \left(\tau^{\mu}_{\phantom{\mu}\nu}e_{\mu}\right)\alpha^{\nu}(v^{\lambda}e_{\lambda})=\tau^{\mu}_{\phantom{\mu}\nu}v^{\lambda}\delta^{\nu}_{\phantom{\nu}\lambda}e_{\mu} $$ which reduces to $\tau^{\mu}_{\phantom{\mu}\lambda}v^{\lambda}e_{\mu}\in V$. The action of the $(1,1)$ tensor $\tau$ on $v$ is denoted, with abuse of notation, as $\tau(v)=(v,\tau)=\tau\cdot v$.

Have a look at this other answers of mine here and here as well.

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  • $\begingroup$ Thanks. Do you really need the notion of basis to show the equivalence in the two notations? I am trying to simplify as much as possible. $\endgroup$ – pluton Oct 12 '15 at 23:33
  • $\begingroup$ One doesn't necessarily need it. You may as well just consider the definition as $\tau\colon V\times V^*\to\mathbb{C}$ and see that $\tau(\cdot, v)\colon V^*\to\mathbb{C}$ and thus isomorphic to a vector (as element of ${V^*}^*$. $\endgroup$ – gented Oct 13 '15 at 7:59

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