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The following question was left as an exercise in my assignment of Manifolds and I am not able to prove this.

Question: Define the map $T^{*} : L^{k}(W) \to L^{k} (V)$ , where $\alpha \in L^{k}(W)$ defined by $T^{*} (\alpha) ( v_1,...,v_k) = \alpha( T(v_1),...,T(v_k))$. $T^{*} (\alpha)$ is called pullback map of $\alpha$ by $T^{*}$.

Here $L^k (V)= V^{*} \oplus ...\oplus V^{*}$.

$T^{*} (\alpha) (v_1+ w_1,...,v_k +w_k)= \alpha ( T(v_1+w_1) ,...,T(v_k+w_k)) = \alpha( T(v_1) +T(w_1), ..., T(v_k) +T(w_k))$.

But I am not able to prove the RHS equal to $\alpha(T(v_1) ,..., T(v_k) )$ + $\alpha( T(w_1,...,T(w_k))$.

Can you please help me with this?

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    The map $T$ itself is linear (as part of the definition, unstated in the question). Does that help? – Andrew D. Hwang May 20 '22 at 12:14
  • @AndrewD.Hwang No, It doesn't help. Can u please elaborate? –  Jun 04 '22 at 09:42
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    My brain is in knots. More details, please. What is $T$? What does $\alpha \in V^* \oplus V^*$ mean, what sort of map is it? – AlvinL Jul 09 '22 at 11:27
  • I think $T^*\alpha$ is supposed to be multilinear. Why else would direct sums be mentioned? Also, a pullback refers to components – AlvinL Jul 09 '22 at 11:34
  • There is no question in your post. Can you state clearly what is to be proved here? – Arctic Char Jul 09 '22 at 11:44
  • @ArcticChar There is . You can have a look at the edit. –  Jul 09 '22 at 11:45
  • Ok. Then of course you are not able to prove it. It is false. – Arctic Char Jul 09 '22 at 11:47
  • e.g. Take $k=2$ and $\alpha (v, w) = v\cdot w$ (the dot product), and say $T = I : V\to V$ is the identity. Then $(v_1+w_1)\cdot (v_2+w_2) \neq v_1\cdot v_2 + w_1\cdot w_2$. – Arctic Char Jul 09 '22 at 11:50

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Suffices to explore $k=2$. This is the only way I can make sense of this picture.

I am assuming $\alpha \in W^*\oplus W^*$ means $\alpha : W\oplus W \to \mathbb K$ is bilinear. This means that $T^*\alpha$ is bilinear map, not linear. So, in general, $\alpha \in L^k(W)$ is a $k$-linear map.

Assume $T:V\to W$ is linear. Then $$ \begin{align*} (T^*\alpha)(\lambda u_1+\mu v_1,u_2) &= \alpha (T(\lambda u_1+\mu v_1), Tu_2) \\ &= \alpha (\lambda Tu_1 + \mu Tv_1, Tu_2) \\ &= \alpha (\lambda Tu_1,Tu_2) + \alpha (\mu Tv_1,Tu_2) \\ &= \lambda\alpha (Tu_1,Tu_2) + \mu\alpha (Tv_1,Tv_2) \\ &= \lambda(T^*\alpha)(u_1,u_2) + \mu(T^*\alpha)(v_1,u_2) \end{align*} $$ and similarly for linearity in second component.

And for linearity of $T^*$ one has $$ \begin{align*} T^*(\lambda\alpha + \mu\beta)\mathbf v &= (\lambda\alpha+\mu\beta)(Tv_1,Tv_2) \\& = \lambda\alpha(Tv_1,Tv_2) + \mu\beta (Tv_1,Tv_2)\\ &= \lambda(T^*\alpha)\mathbf v + \mu(T^*\beta)\mathbf v \\ &= (\lambda T^*\alpha + \mu T^*\beta)\mathbf v \end{align*} $$ for every $(v_1,v_2) =:\mathbf v \in V\oplus V$.

AlvinL
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  • Can you please help with this question also: https://math.stackexchange.com/questions/4443610/every-w-in-omega2-v-is-decomposable-if-operatornamedimv-3 . Bounty is expiring in 2 days. –  Jul 09 '22 at 11:42
  • OK $k$-linear is the better word for it. That's what i meant with 'linear' – Rutger Moody Jul 09 '22 at 12:41